# Maharashtra Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.6

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.6 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.6

Evaluate:

Question 1.
$$\int \frac{2 x+1}{(x+1)(x-2)} d x$$
Solution:
Let I = $$\int \frac{2 x+1}{(x+1)(x-2)} d x$$
Let $$\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}$$
∴ 2x + 1 = A(x – 2) + B(x + 1)
Put x + 1 = 0, i.e. x = -1, we get
2(-1) + 1 = A(-3) + B(0)
∴ A = $$\frac{1}{3}$$
Put x – 2 = 0, i.e. x = 2, we get
2(2) + 1 = A(0) + B(3)
∴ B = $$\frac{5}{3}$$

Question 2.
$$\int \frac{2 x+1}{x(x-1)(x-4)} d x$$
Solution:
Let I = $$\int \frac{2 x+1}{x(x-1)(x-4)} d x$$
Let $$\int \frac{2 x+1}{x(x-1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-4}$$
∴ 2x + 1 = A(x – 1)(x – 4) + Bx(x – 4) + Cx(x – 1)
Put x = 0, we get
2(0) + 1 = A(-1)(-4) + B(0)(-4) + C(0)(-1)
∴ 1 = 4A
∴ A = $$\frac{1}{4}$$
Put x – 1 = 0, i.e. x = 1, we get
2(1) + 1 = A(0)(-3) + B(1)(-3) + C(1)(0)
∴ 3 = -3B
∴ B = -1
Put x – 4 = 0, i.e x = 4, we get
2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(3)
∴ 9 = 12C
∴ C = $$\frac{3}{4}$$

Question 3.
$$\int \frac{x^{2}+x-1}{x^{2}+x-6} d x$$
Solution:

∴ 1 = A(x – 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
1 = A(-5) + B (0)
∴ A = $$\frac{-1}{5}$$
Put x – 2 = 0, i.e. x = 2, we get
1 = A(0) + B(5)
∴ B = $$\frac{1}{5}$$

Question 4.
$$\int \frac{x}{(x-1)^{2}(x+2)} d x$$
Solution:
Let I = $$\int \frac{x}{(x-1)^{2}(x+2)} d x$$
Let $$\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2}$$
∴ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)2
Put x – 1 = 0, i.e. x = 1, we get
1 = A(0)(3) + B(3) + C(0)
∴ B = $$\frac{1}{3}$$
Put x + 2 = 0, i.e. x = -2, we get
-2 = A (-3)(0) + B(0) + C(9)
∴ C = $$-\frac{2}{9}$$
Put x = -1, we get,
-1 = A(-2)(1) + B(1) + C(4)
But B = $$\frac{1}{3}$$ and C = $$-\frac{2}{9}$$

Question 5.
$$\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$$
Solution:
Let I = $$\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$$
Let $$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$$
∴ 3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)2
Put x + 1 = 0, i.e. x = -1, we get
3(-1) – 2 = A(0)(2) + B(2) + C(0)
∴ -5 = 2B
∴ B = $$-\frac{5}{2}$$
Put x + 3 = 0, i.e. x = -3, we get
3(-3) – 2 = A(-2)(0) + B(0) + C(4)
∴ -11 = 4C
∴ C = $$-\frac{11}{4}$$
Put x = 0, we get
3(0) – 2 = A(1)(3) + B(3) + C(1)
∴ -2 = 3A + 3B + C
But B = $$-\frac{5}{2}$$ and C = $$-\frac{11}{4}$$

Question 6.
$$\int \frac{1}{x\left(x^{5}+1\right)} d x$$
Solution:
Let I = $$\int \frac{1}{x\left(x^{5}+1\right)} d x$$
= $$\int \frac{x^{4}}{x^{5}\left(x^{5}+1\right)} d x$$

Question 7.
$$\int \frac{1}{x\left(x^{n}+1\right)} d x$$
Solution:

Question 8.
$$\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x$$
Solution:

Let $$\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$$
∴ 5x2 + 20x + 6 = A(x + 1)2 + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i.e. x = -1, we get
5(1) + 20(-1) + 6 = A(0) + B(-1)(0) + C(-1)
∴ -9 = -C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = -1