Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2526 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{1}{2}, 3,2\right)$
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
$$
\begin{aligned}
& \Rightarrow \lambda=2 \lambda \text { and } 5=4 \lambda+\mu \\
& \therefore \quad=\frac{1}{2} \text { and } \mu=3 \\
&=\frac{1}{2} \int \frac{2 x+4}{x^2+4 x+5} d x+3 \int \frac{d x}{x^2+4 x+5} \\
&=\frac{1}{2} \log \left(x^2+4 x+5\right)+3 \int \frac{d x}{(x+2)^2+(1)^2} \\
&=\frac{1}{2} \log \left(x^2+4 x+5\right)+3 \tan ^{-1}(x+2)+C \\
& \therefore \quad a=\frac{1}{2}, b=3, k=2
\end{aligned}
$$
Put $x+5=\lambda(2 x+4)+\mu$
On comparing both sides, we get
$$
\begin{aligned}
& \Rightarrow \lambda=2 \lambda \text { and } 5=4 \lambda+\mu \\
& \therefore \quad=\frac{1}{2} \text { and } \mu=3 \\
&=\frac{1}{2} \int \frac{2 x+4}{x^2+4 x+5} d x+3 \int \frac{d x}{x^2+4 x+5} \\
&=\frac{1}{2} \log \left(x^2+4 x+5\right)+3 \int \frac{d x}{(x+2)^2+(1)^2} \\
&=\frac{1}{2} \log \left(x^2+4 x+5\right)+3 \tan ^{-1}(x+2)+C \\
& \therefore \quad a=\frac{1}{2}, b=3, k=2
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.