Question

Integrate the following$\int \frac{1}{x^2+8x+20}dx$

• A. $\frac{1}{2}si{n}^{-1}\frac{x+4}{2}+C$
• B. $\frac{1}{2}co{t}^{-1}\frac{x+4}{2}+C$
• C. $\frac{1}{2}ta{n}^{-1}\frac{x+4}{2}+C$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{1}{2}ta{n}^{-1}\frac{x+4}{2}+C$

Solving this, we get

$\int \frac{1}{x^2+8x+20}dx$

we know that $x^2+8x+20=(x^2+2(4)x+4^2)+4=(x+4{)}^2+2^2$

$⟹\int \frac{1}{(x+4{)}^2+2^2}dx$

$⟹\frac{1}{2}{\tan }^{-1}\frac{x+4}{2}+C$