Question

Solve:

$\int \frac{2x+3}{\sqrt{x^2+3x-4}}dx$

• A. $\sqrt{x^2+3x-4}+c$
• B. $2\sqrt{x^2+3x-4}+c$
• C. $2(x^2+3x-4{)}^{3/2}+c$
• D. $-2(x^2+3x-4{)}^{3/2}+c$

Answer 1

The correct option is B $2\sqrt{x^2+3x-4}+c$

1. $\int x^n=\frac{x^n+1}{n+1}+C$
Given,
$\int \frac{2x+3}{\sqrt{x^2+3x-4}}dx\to \left(1\right)$
Let, $t=x^2+3x-4$
Differentiating t w.r.t. $x$,
$\frac{dt}{dx}=2x+3$
$\Rightarrow dt=(2x+3)dx$
Substituting these values in expression 1,
$\int \frac{(2x+3)dx}{\sqrt{x^2+3x-4}}=\int \frac{dt}{\sqrt{t}},$
$=\int \frac{1}{t^{\frac{1}{2}}}dt,$
$=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c,$ (Using Ref 1)
$=2\sqrt{t}+c$
Substituting the value of t,
$\int \frac{2x+3}{\sqrt{x^2+3x-4}}dx=2\sqrt{x^2+3x-4}+c$