Question

Solve: $\int \frac{dx}{\sqrt{21-4x-x^2}}$

• A. $\arcsin \frac{(x+4)}{5}+C$
• B. $\arcsin \frac{(x+2)}{5}+C$
• C. $\arcsin \frac{(x-4)}{5}+C$
• D. None of these

Answer 1

The correct option is B $\arcsin \frac{(x+2)}{5}+C$

Let $I=\int \frac{dx}{\sqrt{21-4x-x^2}}$

$\int \csc u\cot udu=-\csc u+C$

$\int \frac{du}{\sqrt{a^2-u^2}}=\text{arcsin}\frac{u}{a}+C$

$\int \frac{du}{a^2+u^2}=\frac{1}{a}\text{arctan}\frac{u}{a}+C$

$\int \frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}\text{arcsec}\frac{u}{a}+C$

$\int \frac{du}{u^2-a^2}=\frac{1}{2a}\ln \left|\frac{u-a}{u+a}\right|+C$

$\int \frac{du}{a^2-u^2}=\frac{1}{2a}\ln \left|\frac{a+u}{a-u}\right|+C$

Taking only the denominator part, split $21$ as $25-4$

Now this $4$ goes with the other terms to make another term $u^2$, where $u=x+2$.

$\Rightarrow I=\int \frac{dx}{\sqrt{5^2-(x+2{)}^2}}=\text{arcsin}\frac{(x+2)}{5}+C$