Integral Calculus
86339
If
\(\int \frac{x+5}{x^{2}+4 x+5} d x=a \log \left(x^{2}+4 x+5\right)\)
\(+\operatorname{btan}^{-1}(x+k)+C\), then \((a, b, k)\) equals
1 \(\left(\frac{1}{2}, 3,2\right)\)
2 \(\left(\frac{1}{2}, 1,2\right)\)
3 \(\left(\frac{1}{2}, 3,1\right)\)
4 \((1,3,2)\)
Explanation:
(A) : Given,
\(\int \frac{x+5}{x^{2}+4 x+5} d x=a \log \left(x^{2}+4 x+5\right)+b \tan ^{-1}(x+k)+C\)
L.H.S, \(\quad I=\int \frac{x+5}{x^{2}+4 x+5} d x\)
\(=\int \frac{x+2+3}{x^{2}+4 x+5} d x=\int \frac{x+2}{x^{2}+4 x+5} d x+\int \frac{3}{x^{2}+4 x+5} d x\)
Let,
\(x^{2}+4 x+5=t\)
\((2 \mathrm{x}+4) \mathrm{dx}=\mathrm{dt}\)
\((\mathrm{x}+2) \mathrm{dx}=\frac{\mathrm{dt}}{2}\)
\(\therefore \quad \mathrm{I}=\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}+\int \frac{3 \mathrm{dx}}{(\mathrm{x}+2)^{2}+(1)^{2}}\)
\(=\frac{1}{2} \log |\mathrm{t}|+3 \times \frac{1}{1} \tan ^{-1}\left(\frac{\mathrm{x}+2}{1}\right)+\mathrm{C}\)
\(=\frac{1}{2} \log \left(\mathrm{x}^{2}+4 \mathrm{x}+5\right)+3 \tan ^{-1}(\mathrm{x}+2)+\mathrm{C}\)
On comparing with right hand side we get -
\(\mathrm{a}=\frac{1}{2}, \quad \mathrm{~b}=3, \quad \mathrm{k}=2\)
So, \((\mathrm{a}, \mathrm{b}, \mathrm{k})=\left(\frac{1}{2}, 3,2\right)\)
