Q93 / 160 · TS EAMCET 2015 · Mathematics
If $\int \frac{x+5}{x^2+4 x+5} d x=a \log \left(x^2+4 x+5\right)$ $+b \tan ^{-1}(x+k)+C$, then $(a, b, k)$ equals
- Correct answerA$\left(\frac{1}{2}, 3,2\right)$
- B$\left(\frac{1}{2}, 1,2\right)$
- C$\left(\frac{1}{2}, 3,1\right)$
- D$(1,3,2)$
Solution
Let $I=\int \frac{x+5}{x^2+4 x+5} d x$
Put $x+5=\lambda(2 x+4)+\mu$
On comparing both sides, we get
$$
\begin{aligned}
& \Rightarrow \lambda=2 \lambda \text { and } 5=4 \lambda+\mu \\
& \therefore \quad=\frac{1}{2} \text { and } \mu=3 \\
&=\frac{1}{2} \int \frac{2 x+4}{x^2+4 x+5} d x+3 \int \frac{d x}{x^2+4 x+5} \\
&=\frac{1}{2} \log \left(x^2+4 x+5\right)+3 \int \frac{d x}{(x+2)^2+(1)^2} \\
&=\frac{1}{2} \log \left(x^2+4 x+5\right)+3 \tan ^{-1}(x+2)+C \\
& \therefore \quad a=\frac{1}{2}, b=3, k=2
\end{aligned}
$$
Hence, the correct answer is option: 'A'.