Question

Find the integral $\int \frac{x+1}{x^2+2x-4}dx$

• A. $\ln (x^2+2x-4)+C$
• B. $\frac{1}{2}\ln (x^2+2x)+C$
• C. $\frac{1}{2}\ln (x^2+2x-4)+C$
• D. $\frac{1}{2}\ln (x+1)+C$

Answer 1

The correct option is C $\frac{1}{2}\ln (x^2+2x-4)+C$

Given, $f\left(x\right)=\frac{x+1}{x^2+2x-4}$
So, to integrate the given function we use substitution method.
Let $t=x^2+2x-4\Rightarrow \frac{dt}{dx}=2x+2=2(x+1)$
Or, $\frac{dt}{2}=(x+1)dx$
Thus, substituting this value in the given function we have,
$\int \frac{x+1}{x^2+2x-4}dx=\int \frac{1}{2t}dt=\frac{1}{2}\ln t+C$
$\Rightarrow \frac{1}{2}\ln (x^2+2x-4)+C$