# Get Answers to all your Questions

### Answers (1)

Answer:

$\frac{-1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x+1|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{1+x+x^{2}+x^{3}} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{1+x+x^{2}+x^{3}} d x \\ &I=\int \frac{1}{(1+x)+x^{2}(x+1)} d x \\ &I=\int \frac{d x}{(1+x)\left(1+x^{2}\right)} \end{aligned}

\begin{aligned} &\frac{1}{\left(1+x^{2}\right)(1+x)}=\frac{A x+B}{1+x^{2}}+\frac{C}{1+x} \\ &1=(A x+B)(1+x)+C\left(1+x^{2}\right) \\ &1=x^{2}(A+C)+x(B+A)+(B+C) \end{aligned}

Equating similar terms

\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}                       (1)

\begin{aligned} &B+A=0 \\ &A=-B \end{aligned}                       (2)

\begin{aligned} &B+C=1 \\ &-A-A=1 \\ &-2 A=1 \\ &A=\frac{-1}{2} \end{aligned}

Equation (1)

$C=\frac{1}{2}$

Equation (2)

$B=\frac{1}{2}$
\begin{aligned} &\frac{1}{1+x+x^{2}+x^{3}}=\frac{-\frac{1}{2} x+\frac{1}{2}}{x^{2}+1}+\frac{\frac{1}{2}}{x+1} \\ &=\frac{-x+1}{2\left(x^{2}+1\right)}+\frac{1}{2(x+1)} \end{aligned}
\begin{aligned} &I=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x+1} d x \\ &I=\frac{-1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x+1|+C \end{aligned}

View full answer

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support