Inter 2nd Year Maths Exercise 7e Solutions

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7e will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7e

Integrals Exercise 7e Solutions

I. Integrate the Rational Functions.

Question 1.
\(\frac{x}{(x+1)(x+2)}\)
Solution:
∫\(\frac{x}{(x+1)(x+2)}\) dx
Let \(\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}\)
= \(\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}+2)}\)
⇒ \(\frac{x}{(x+1)(x+2)}\) = \(\frac{\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)}{(\mathrm{x}+1)(\mathrm{x}+2)}\)
⇒ x = x(A + B) + 2A + B
On equating the coefficient of x and constant terms on both sides, we get
A + B = 1 ………. (1) & 2A + B = 0 ………….. (2)
(2) – (1) ⇒ A = -1
Put the value of A in eq(1), we get
-1 + B = 1 ⇒ B = 2
∴ ∫\(\frac{x}{(x+1)(x+2)}\) dx = ∫\(\frac{-1}{(x+1)}\) dx + ∫\(\frac{2}{(x+2)}\) dx
= log(x + 2)² – log(x + 1) + C [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= \(\log \frac{(x+2)^2}{|x+1|}+C\) [∵ loga – logb = log \(\frac{\mathrm{b}}{\mathrm{a}}\)]

Question 2.
\(\frac{1}{x^2-9}\)
Solution:
∫\(\frac{1}{x^2-9}\) dx
= ∫\(\frac{1}{x^2-3^2}\) dx = ∫\(\frac{1}{(x+3)(x-3)}\) dx
Let \(\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}\)
⇒ 1 = A(x – 3) + B(x + 3) ⇒ 1 = x(A + B) + (-3A + 3B)
On equating the coefficient of x and constant terms on both sides, we get
A + B = 0 and -3A + 3B = 1
On solving, we get A = –\(\frac{1}{6}\) and B = \(\frac{1}{6}\)
∴ ∫\(\frac{1}{(x+3)(x-3)}\) dx = \(\frac{-1}{6(x+3)}\) dx + \(\frac{1}{6(x-3)}\) dx
= \(-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+C\)
= \(\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+C\) [∵ loga – logb = log \(\frac{\mathrm{b}}{\mathrm{a}}\)]

Question 3.
\(\frac{1-x^2}{x(1-2 x)}\)
Solution:
Let ∫\(\frac{1-x^2}{x(1-2 x)}\) dx
Here, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.

On comparing the coefficient of x and constant terms on both sides, we get
2A + B = \(\frac{1}{2}\) and -A = -1 ⇒ A = 1
⇒ 2 × 1 + B = \(\frac{1}{2}\) ⇒ B = \(\frac{1}{2}\) – 2 = \(\frac{-3}{2}\)
∴ I₂ = ∫\(\left[\frac{1}{x}-\frac{3}{2(2 x-1)}\right]\) dx = ∫\(\frac{1}{x} d x-\frac{3}{2} \int \frac{1}{2 x-1}\) dx
⇒ I₂ = log x – \(\frac{3}{2} \frac{\log |2 x-1|}{2}\) + C₂
[∵ ∫\(\frac{1}{x}\) dx = log x + C]
Thus, on putting the values of I₁ and I₂ in eq(1), we get
I = \(\frac{1}{2}\)x + log x – \(\frac{3}{4}\) log|2x – 1| + C [∵ C₁ + C₂ = C]

II. Integrate the Rational Functions.

Question 1.
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
Solution:
Let \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\)
⇒ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
⇒ \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
⇒ 3x – 1 = A[x² – 5x + 6] + B[x² – 4x + 3] + C[x² – 3x + 2]
⇒ 3x – 1 = x²(A + B + C) + x(-5A – 4B – 3C) + (6A + 3B + 2C)
On equating the coefficients of x², x and constant terms on both sides, we get
A + B + C = O
-5A – 4B – 3C = 3 ………… (ii)
6A + 3B + 2C = -1 ……….. (iii)
From eq(i), we get A = – (B + C)
On putting the value of A in eq (ii) and (iii), we get
-5{-(B + C)} – 4B – 3C = 3
⇒ 5B + 5C – 4B – 3C = 3 = B + 2C = 3 ……….. (iv)
and 6{-(B + C)} + 3B + 2C = -1
⇒ -6B – 6C + 3B + 2C = -1 – 3B – 4C = -1 ………. (v)
On solving eqs (iv) and (v), we get C = 4
On putting the value of C in eq(iv), we get
B + 2 × 4 = 3 ⇒ B = -5
Putting the value of B and C in eq(i), we get
A +(-5) + 4 = 0 ⇒ A = 1
∴ A = 1, B = -5, C = 4
Now, ∫\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= ∫\(\left(\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\right)\) dx
= ∫\(\frac{1}{(x-1)}\) dx + ∫\(\frac{(-5)}{(x-2)}\) dx + ∫\(\frac{4}{x-3)}\) dx
= log|x – 1| – 5 log|x – 2| + 4 log|x – 3| + C

Question 2.
\(\frac{x}{(x-1)(x-2)(x-3)}\)
Solution:
∫\(\frac{x}{(x-1)(x-2)(x-3)}\) dx
Let \(\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\) = \(\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\)
⇒ \(\frac{x}{(x-1)(x-2)(x-3)}\) = \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
⇒ x = A[x² – 5x + 6] + B[x² – 4x + 3] + C[x² – 3x + 2]
⇒ x = x² (A + B + C) + x(-5A – 4B – 3C)+ (6A + 3B + 2C)
On comparing the coefficients of x², x and constant terms on both sides, we get
A + B + C = 0 ………… (1)
-5A – 4B – 3C = 1 …………… (ii)
6A + 3B + 2C = O ………. (iii)
From eq(i), we get A = – (B + C)
On putting the value of A in eqs (ii) and (iii), we get
-5{-(B + C)} – 4B – 3C = 1
5B + 5C – 4B – 3C = 1 = B + 2C = 1 ……….. (iv)
and 6 (-(B + C)} + 3B + 2C = 0
⇒ -3B – 4C = 0 ……………. (v)
On solving (iv) & (v), we get C = \(\frac{3}{2}\)
On putting the value of C in eq(iv), we get
B + 2(\(\frac{3}{2}\)) = 1 ⇒ B = -2
Now, put the values of B and C in eq (1) we get
A – 2 + \(\frac{3}{2}\) = 0 ⇒ A = 2 – \(\frac{3}{2}\) = \(\frac{1}{2}\)
∴ A = \(\frac{1}{2}\), B = -2 and C = \(\frac{3}{2}\)
Now, ∫\(\frac{x}{(x-1)(x-2)(x-3)}\) dx [∵ ∫\(\frac{1}{x}\) dx = log|x| + C]
= ∫\(\frac{\mathrm{A}}{(\mathrm{x}-1)}\) dx + ∫\(\frac{\mathrm{B}}{(\mathrm{x}-2)}\) dx + ∫\(\frac{\mathrm{C}}{(\mathrm{x}-3)}\) dx
= \(\frac{1}{2}\) ∫\(\frac{1}{x-1)}\) dx – 2 ∫\(\frac{1}{(x-2)}\) dx + \(\frac{3}{2}\) ∫\(\frac{1}{(x-3)}\) dx
= \(\frac{1}{2}\) log|x – 1| – 2 log|x – 2| + \(\frac{3}{2}\) log|x – 3| + C

Question 3.
\(\frac{2 x}{x^2+3 x+2}\)
Solution:
Let \(\frac{2 x}{x^2+3 x+2}=\frac{2 x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\)
⇒ \(\frac{2 \mathrm{x}}{(\mathrm{x}+2)(\mathrm{x}+1)}\) = \(\frac{\mathrm{A}(\mathrm{x}+1)+\mathrm{B}(\mathrm{x}+2)}{(\mathrm{x}+2)(\mathrm{x}+1)}\)
⇒ 2x = x(A + B) + (A + 2B)
On comparing the coefficient of x and constant terms, on both sides, we get
A + B = 2 …………. (1)
and A + 2B = 0 ………….. (ii)
(i) – (ii) ⇒ B = 2 ⇒ B = -2
On putting the value of B in eq (i), we get
A – 2 = 2 ⇒ A = 4
∴ ∫\(\frac{2 \mathrm{x}}{(\mathrm{x}+2)(\mathrm{x}+1)}\) dx = ∫\(\frac{\mathrm{A}}{(\mathrm{x}+2)}\) dx + ∫\(\frac{\mathrm{B}}{(\mathrm{x}+1)}\) dx
= ∫\(\frac{4}{(x+2)}\) dx + ∫\(\frac{(-2)}{(x+1)}\) dx
= 4log |x + 2| – 2log |x + 1| + C

Question 4.
\(\frac{x}{\left(x^2+1\right)(x-1)}\)
Solution:
∫\(\frac{x}{\left(x^2+1\right)(x-1)}\) dx
Let \(\frac{x}{\left(x^2+1\right)(x-1)}\) = \(\frac{A}{x-1}+\frac{B x+C}{x^2+1}\)
⇒ x = A(x² + 1) + (Bx + C) (x – 1) ………….(1)
Substituting x = 1 and O in eq (ii), we get
1 = A(2) and 0 = A – C ⇒ A = \(\frac{1}{2}\) and C = A = \(\frac{1}{2}\)
On equating the coefficient of x² on the both sides in eq (ii), we get
0 = A + B ⇒ B = -A = \(\frac{-1}{2}\)

Question 5.
\(\frac{2}{(1-x)\left(1+x^2\right)}\)
Solution:
Let \(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{A}{1-x}+\frac{B x+C}{1+x^2}\)
⇒ \(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{A\left(1+x^2\right)+(B x+C)(1-x)}{(1-x)\left(1+x^2\right)}\)
⇒ A + Ax+ Bx+ C – Bx² – Cx
⇒ 2 = x²(A – B) + x(B – C) + (A + C)
On comparing the coefficients of x², x and constant terms on both sides, we get
A -B = 0 ⇒ A = B ………… (i)
B – C = 0 ⇒ B = C …………. (ii)
and A + C = 2 ………… (iii)
From eq (i) and (ii), we get A = C put this value in eq (iii) we get 2A = 2 ⇒ A = 1
Put the value of A in eq (ii) and (iii), we get B = 1 and C = 1
∴ \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}\)
= ∫\(\frac{1}{1-x}\) dx + \(\frac{1}{2}\) ∫\(\frac{2 x}{x^2+1}\) dx + ∫\(\frac{1}{x^2+1}\) dx
Let x² + 1 = t ⇒ 2x dx = dt
∫\(\frac{2 x}{x^2+1}\) dx = ∫\(\frac{1}{t}\) dt = log t
= log |1 – x| + \(\frac{1}{2}\) log(1 + x²) + tan^-1x +

Question 6.
\(\frac{3 x-1}{(x+2)^2}\)
Solution:
Let \(\frac{3 x-1}{(x+2)^2}\) = \(\frac{A}{x+2}+\frac{B}{(x+2)^2}\)
⇒ 3x – 1 = A(x + 2) + B
On equating the coefficient of x and constant terms on both sides, we get A = 3 and
2A + B = -1 ⇒ 2(3) + B = -1 ⇒ B = -7
∴ \(\frac{3 x-1}{(x+2)^2}\) = \(\frac{3}{x+2}+\frac{7}{(x+2)^2}\)
∴ ∫\(\frac{3 x-1}{(x+2)^2}\) = 3 ∫\(\frac{1}{(x+2)}\) dx – 7 ∫\(\frac{1}{(x+2)^2}\) dx
= 3 log |x + 2| – 7 \(\left(\frac{-1}{x+2}\right)\) + C
= 3 log |x + 2| + \(\frac{7}{x+2}\) + C

Question 7.
\(\frac{1}{x\left(x^n+1\right)}\)
[Hint: Multiply numerator and denominator by x^n-1 and put xⁿ = t]
Solution:
Let I = ∫\(\frac{1}{x\left(x^n+1\right)}\) dx
Put xⁿ = t ⇒ nx^n-1dx = dt ⇒ x^n-1 dx = \(\frac{1}{n}\) dt
∴ ∫\(\frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{x}^{\mathrm{n}}\left(\mathrm{x}^{\mathrm{n}}+1\right)}\) dx = \(\frac{1}{n} \int \frac{1}{t(t+1)} d t\) ………….. (i)
Now, \(\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}\)
1 = A(1 + t) + Bt ………….. (ii)
On substituting t = 0, -1 in equation (ii), we get
A = 1 and B = -1
∴ \(\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(t+1)}\)
∴ \(\frac{1}{n} \int\left[\frac{1}{t}-\frac{1}{(t+1)}\right]\) dt = \(\frac{1}{n}\) [log |t| – log |t + 1| + C
= \(\frac{1}{n}\) [log |xⁿ| = log(xⁿ + 1)] + C [put t = xⁿ]
= \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|\) + C

Question 8.
\(\frac{1}{\left(e^x-1\right)}\) [Hint: Put e^x = t]
Solution:
Let I = ∫\(\frac{1}{\left(e^x-1\right)}\) dx
On mutliplying numerator & denominator by e^-x,
I = ∫\(\frac{e^{-x}}{1-e^{-x}}\) dx
Put 1 – e^-x = t ⇒ -e^-x (-1) = \(\frac{d t}{d x}\) ⇒ e^-x dx = dt
∴ I = ∫\(\frac{d t}{t}\) = log |t| + C = log |1 – e^-x| + C
= \(\log \left|\frac{e^x-1}{e^x}\right|\) + C

Question 9.
\(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\)
Solution:
Let I = ∫\(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\) dx
Put e^x = t ⇒ e^x dx = dt
∴ I = ∫\(\frac{e^x}{1+t)(2+t)} \frac{d t}{e^x}\) = ∫\(\frac{1}{(1+t)(2+t)}\) dt
Let \(\frac{1}{(1+t)(2+t)}\) = \(\frac{A}{(1+t)}+\frac{B}{(2+t)}\) = \(\frac{A(2+t)+B(1+t)}{(1+t)(2+t)}\)
⇒ 1 = (A + B)t + (2A + B)
On equating the coefficients of t and constant terms on both sides, we get A + B = O and 2A + B = 1
On solving both equations we get A = 1 & B = -1
I = ∫\(\frac{1}{(1+t)}\) dt – ∫\(\frac{1}{(2+t)}\) dt
= log |1 + t| – log |2 + t| + C
= \(\log \left|\frac{1+t}{2+t}\right|\) + C
= \(\log \left|\frac{1+e^x}{2+e^x}\right|\) + C

Question 10.
\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\)
Solution:
∫\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) dx
Let \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) = \(\frac{A x+B}{x^2+1}\) + \(\frac{C x+D}{x^2+1}\)
⇒ 1 = (Ax + B) (x² + 4) + (Cx + D) (x² + 1)
On comparing the coefficients of x³, x², x and constant terms on both sides, we get
A + C = 0, B + D = 0, 4A + C = 0 and 4B + D = 1
On solving these equations, we get
A = 0, C = OB = and D = \(\frac{-1}{3}\)
∴ ∫\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) dx = \(\frac{1}{3}\left(\int \frac{1}{\left(x^2+1\right)}-\frac{1}{\left(x^2+4\right)}\right)\) dx
= \(\frac{1}{3}\left[\tan ^{-1} x-\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]\) + C

III. Integrate the Following Rational Functions.

Question 1.
\(\frac{x}{(x-1)^2(x+2)}\)
Solution:
Let ∫\(\frac{\mathrm{x}}{(\mathrm{x}-1)^2(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{(\mathrm{x}-1)^2}+\frac{\mathrm{C}}{\mathrm{x}+2}\) ………… (1)
x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² ………….. (2)
Put x = 1, -2 in eq (2) we get
1 = B(1 + 2) ⇒ B = 1/3
-2 = C(-2 – 1)² ⇒ C = \(\frac{2}{9}\)
On comparing coefficient of x² in eqn (2) then
0 = A + C ⇒ A = -C = \(\frac{2}{9}\) ⇒ A = \(\frac{2}{9}\)
Substituting A, B & C values in eqn (1) then

Question 2.
\(\frac{3 x+5}{x^3-x^2-x+1}\)
Solution:
x³ – x² – x + 1 = (x – 1)²(x + 1)
Let \(\frac{3 x+5}{x^3-x^2-x+1}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}\) …………. (1)
3x + 5 = A(x² – 1) + B(x + 1) + C(x – 1)²
3x + 5 = x²(A + C) + x (B – 2C) + (B + C – A) ……….. (2)
Put x = 1 then 8 = 2B ⇒ B = 4
Put x = -1 then 2 = 4C ⇒ C = 1/2
On comparing coefficients of x² in eqn (2) then
A + C = O ⇒ A = -C = -1/2
Substitute A, B, C values in eqn(1) then

Question 3.
\(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\)
Solution:
Let \(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\) = \(\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2 x+3}\) ………… (1)
2x – 3 = A(2x + 3)(x + 1) + B(x – 1)(2x + 3) + C(x – 1)(x + 1) ………… (2)
Put x = 1 in eqn (2) then -1 = 10A ⇒ A = \(\frac{-1}{10}\)
Put x = -1 in eqn(2) then -5 = B(-2) ⇒ B = \(\frac{5}{2}\)
Put x = \(\frac{-3}{2}\) in eqn(2) then -6 = C(\(\frac{-3}{2}\) -1)(\(\frac{-3}{2}\) + 1)
-6 = C(\(\frac{-5}{2}\))(\(\frac{-1}{2}\)) ⇒ -24 = 5C ⇒ C = –\(\frac{24}{5}\)
Substitute A, B, C values in eqn(1) then

Question 4.
\(\frac{5 x}{(x+1)\left(x^2-4\right)}\)
Solution:
Let \(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\) …….. (1)
5x = A(x² – 4) + B(x + 1)(x – 2) + C(x + 1)(x + 2) ………… (2)
Put x = -1 in eqn(2) then -5 ⇒ -3A = -5 ⇒ A = \(\frac{5}{3}\)
Put x = 2 in eqn(2) then 10 = C(3)(4) ⇒ C = \(\frac{5}{6}\)
Put x = -2 in eqn(2) then -10 = B(-4)(-1) ⇒ B = –\(\frac{5}{2}\)
Substitute A, B, C values in eqn(1) then

Question 5.
\(\frac{x^3+x+1}{x^2-1}\)
Solution:
Let \(\frac{x^3+x+1}{x^2-1}\) = x + \(\frac{2 x+1}{x^2-1}\) ………….. (1)
Consider \(\frac{2 x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}\) ……….. (2)
2x + 1 = A(x + 1) + B(x – 1) …………. (3)
Put x = 1 in eqn(3) then 3 = 2A ⇒ A = \(\frac{3}{2}\)
Put x = -1 in eqn(3) then -1 = -2B ⇒ D = \(\frac{1}{2}\)
Substitute A, B values in eq(2) then
\(\frac{2 x+1}{x^2-1}\) = \(\frac{3 / 2}{x-1}+\frac{1 / 2}{x+1}\) ……….. (4)
From 1 & 4 we get
\(\frac{x^3+x+1}{x^2-1}\) = x + \(\frac{3}{2(x-1)}\) + \(\frac{1}{2(x+1)}\)
∴ ∫\(\frac{x^3+x+1}{x^2-1}\) dx = v x dx + \(\frac{3}{2}\) ∫\(\frac{1}{x-1}\) dx + \(\frac{1}{2}\) ∫\(\frac{1}{x+1}\) dx
= \(\frac{x^2}{2}\) + \(\frac{3}{2}\) log|x – 1| + \(\frac{1}{2}\)log|x + 1| + C

Question 6.
\(\frac{1}{x^4-1}\)
Solution:
Consider \(\frac{1}{x^4-1}\) = \(\frac{1}{(x-1)(x+1)\left(x^2+1\right)}\)
Let \(\frac{1}{(x+1)(x-1)\left(x^2+1\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^2+1\right)}\) ………… (1)
⇒ 1 = A(x + 1) (x² + 1) + B(x – 1) (x² + 1) + (Cx + D)(x² – 1)
Put x = -1 then 1 = -4B ⇒ B=—1/4
Put x = 1 then 1 = 4A ⇒ A = 1/4
Put x = 0 then 1 = A – B – D
D = \(\frac{1}{4}\) + \(\frac{1}{4}\) – 1 = -2/4 = -1/2
On comparing the coefficients of x on both sides,
0 = A + B – C ⇒ C = A + B = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 0
Substituting the values of A, B, C & D in (1),
\(\frac{1}{(x-1)(x+1)\left(x^2+1\right)}\) = \(\frac{1 / 4}{(x-1)}+\frac{-1 / 4}{x+1}+\frac{-1 / 2}{x^2+1}\)
∴ ∫\(\frac{1}{x^4-1}\) dx = \(\frac{1}{4}\) ∫\(\frac{1}{x-1}\) – \(\frac{1}{4}\) ∫\(\frac{1}{x+1}\) dx – \(\frac{1}{2}\) ∫\(\frac{1}{x^2+1}\) dx
= \(\frac{1}{4}\) log |x – 1| – \(\frac{1}{4}\) log|x + 1| – \(\frac{1}{2}\) tan^-1 + C
= \(\frac{1}{4}\) log\(\frac{x-1}{x+1}\) – \(\frac{1}{2}\)tan^-1x + C [∵ ∫\(\frac{1}{x^2+1}\) dx = tan^-1x + C]

Question 7.
\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) [Hint: Put sin x = t]
Solution:
∫\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) dx
Put sin x = t ⇒ cos xdx = dt
Consider \(\frac{1}{(1-t)(2-t)}\) = \(\frac{A}{1-t}+\frac{B}{2-t}\) ………… (1)
1 = A(2 – t) + B(1 – t)
Put t = 1, 2 then A = 1 & B = -1
Substituting A, B values in eq(1) then

Question 8.
\(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}\)
Solution:

Question 9.
\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)
Solution:
Consider ∫\(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\) dx
Put x² = t ⇒ 2x dx = dt
Let \(\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}\)
⇒ 1 = A(t + 3) + B(t + 1)
Put t = -1 then 2A = 1 ⇒ A = 1/2
Put t = -3 then -2B = 1 ⇒ B = -1/2
∴ ∫\(\frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}+3)}\) = \(\int \frac{\frac{1}{2}}{t+1} d t-\int \frac{1 / 2}{t+3} d t\)
= \(\frac{1}{2}\) ∫\(\frac{1}{t+1}\) dt – \(\frac{1}{2}\) ∫\(\frac{1}{t+3}\) dt
= \(\frac{1}{2}\) log|t + 1| – \(\frac{1}{2}\) log|t + 3| + C
= \(\frac{1}{2}\) log|x² + 1| – \(\frac{1}{2}\) log|x² + 3| + C
= \(\frac{1}{2}\) log\(\left|\frac{x^2+1}{x^2+3}\right|\) + C

Question 10.
\(\frac{1}{x\left(x^4-1\right)}\)
Solution:

Question 11.
\(\frac{1}{x-x^3}\)
Solution:
\(\int \frac{1}{x-x^3} d x\) = \(\int \frac{1}{x\left(1-x^2\right)} d x=\int \frac{1}{x(1-x)(1+x)} d x\)
Consider \(\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}\) ……….. (1)
1 = A(1 – x²) + Bx(1 + x) + Cx(1 – x) ………….. (2)
Put x = 0 in eq (2) then 1 = A
Put x = 1 in eq (2) then 1 = 2B ⇒ B = 1/2
Put x = -1 in eq (2) then 1 = -2C ⇒ C = -1/2
Substitute A, B, C values in eqn(1) then
\(\frac{1}{x\left(1-x^2\right)}\) = \(\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}\)
∫\(\frac{1}{x\left(1-x^2\right)}\) dx = ∫\(\frac{1}{x}\) dx + \(\frac{1}{2}\) ∫\(\frac{1}{1-x}\) dx – \(\frac{1}{2}\) ∫\(\frac{1}{1+x}\) dx
= log x – \(\frac{1}{2}\) log|1 – x| – \(\frac{1}{2}\) log|1 + x| + C
= log x² – \(\frac{1}{2}\) [log(1 – x²)] + C
= \(\frac{1}{2}\) log \(\frac{x^2}{1-x^2}\) + C

Question 12.
\(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\) [Hint: \(\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}\), put x = t⁶]
Solution:

Question 13.
\(\frac{5 x}{(x+1)\left(x^2+9\right)}\)
Solution:
∫\(\frac{5 x}{(x+1)\left(x^2+9\right)}\) dx
Consider \(\frac{5 x}{(x+1)\left(x^2+9\right)}\) = \(\frac{A}{x+1}+\frac{B x+C}{x^2+9}\) ………… (1)
5x = A(x² + 9) + (Bx + C) (x + 1) ……….. (2)
Put x = -1 in eqn(2) then 5(-1) ⇒ 10A = -5 ⇒ A = -1/2
Comparing the coefficient of x² on both sides,
0 = A + B ⇒ B = -A = 1/2
Comparing the coefficient of x² on both sides,
5 = B + C ⇒ 5 = \(-\frac{1}{2}\) + C ⇒ C = 5 – \(-\frac{1}{2}\) = \(-\frac{9}{2}\)
Substituting A, B, C values in eqn (1) then

Question 14.
\(\frac{x^2+x+1}{(x+1)^2(x+2)}\)
Solution:
∫\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) dx
Consider \(\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+2)}\) ……….. (1)
x² + x + 1 = A(x + 1) (x + 2) + B(x + 2) + C(x + 1)² ………….. (2)
Put x = -1 in eq(2) then 1 – 1 + 1 = B ⇒ B = 1
Put x = -2 in eq(2) then 4 – 2 + 1 = C(-1)² ⇒ C = 3
Comparing the coefficients of x² on both sides,
1 = A + C ⇒ A = 1 – C = 1 – 3 = -2
Substituting A, B, C values in eq (1) then
\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) = \(\frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}\)
∫\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) dx
= – ∫\(\frac{1}{x+1}\) dx + ∫\(\frac{1}{(x+1)^2}\) dx + 3∫\(\frac{1}{x+2}\) dx
= -2 log|x + 1| – \(\frac{1}{x+1}\) + 3 log|x + 2| + C