Let $$I=\int \frac{3 x-2}{(x+3)(x+1)^2} d x$$

By partial fraction

$$\begin{aligned} & \frac{(3 x-2)}{(x+3)(x+1)^2}=\frac{A}{x+3}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2} \quad\text{....(i)} \\ \Rightarrow \quad & (3 x-2)=A(x+1)^2+B(x+3)(x+1)+C(x+3) \\ & =A\left(x^2+1+2 x\right)+B\left(x^2+4 x+3\right)+C(x+3) \\ & (A+B) x^2+(2 A+4 B+C) x+(A+3 B+3 C)\end{aligned}$$

On comparing the like powers of $$x$$, we get

$$\begin{aligned} & A+B=0 \quad \text{.... (ii)}\\ & 2 A+4 B+C=3 \quad \text{.... (iii)}\\ & \text { and } \quad A+3 B+3 C=-2 \quad \text{.... (iv)}\\ \end{aligned}$$

On multiplying Eq. (iii) by 3 and then subtracting it from Eq. (iv), we get

VITEEE 2021 Mathematics - Indefinite Integration Question 1 English Explanation

From Eqs. (ii) and (v), we get

$$B=\frac{11}{4} \text { and } A=\frac{-11}{4}$$

From Eq. (iii),

$$\begin{aligned} C & =3-2 A-4 B \\ & =3-2\left(\frac{-11}{4}\right)-4\left(\frac{11}{4}\right) \\ & =3+\frac{22}{4}-\frac{44}{4}=3-\frac{22}{4}=3-\frac{11}{2} \\ & =-\frac{5}{2} \end{aligned}$$

$$\therefore$$ From Eq. (i), we get

$$\begin{aligned} & \frac{(3 x-2)}{(x+3)(x+1)^2}=\frac{-11}{4(x+3)}+\frac{11}{4(x+1)} \cdot \frac{-5}{2(x+1)^2} \\ \Rightarrow & \int \frac{(3 x-2)}{(x+3)(x+1)^2} d x \\ = & \int\left\{\frac{-11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}\right\} d x \\ = & -\frac{11}{4} \log |x+3|+\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}+C \\ = & \frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C \end{aligned}$$

Algebra

Trigonometry

Calculus

Coordinate Geometry

Indefinite Integration · Mathematics · VITEEE

MCQ (Single Correct Answer)