KCET · Maths · Indefinite Integration
If \(\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}\), then \(\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C\) is equal to
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{6}\)
- C 0
- D \(\frac{5 \pi}{6}\)
Answer & Solution
Correct Answer
(D) \(\frac{5 \pi}{6}\)
Step-by-step Solution
Detailed explanation
Given,
\(\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}...(i)\)
\(\begin{aligned} \Rightarrow \quad(x+1)^{2} &=A\left(x^{2}+1\right)+(B x+c)(x) \\ x^{2}+1+2 x &=A x^{2}+A+B x^{2}+C x \\ &=(A+B) x^{2}+C x+A \end{aligned}\)
On comparing the coefficient of like powers on both sides, we get
\(A+B=1...(ii)\)
and \(\quad C=2, \quad A=1\)
From Eq. (ii), \(B=0\)
Then, \(\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C\)
\(=\sin ^{-1}(1)+\tan ^{-1}(0)+\sec ^{-1}(2)\)
\(=\sin ^{-1} \sin \frac{\pi}{2}+\tan ^{-1} \tan 0+\sec ^{-1} \sec \frac{\pi}{3}\)
\(=\frac{\pi}{2}+0+\frac{\pi}{3}=\frac{5 \pi}{6}\)
\(\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}...(i)\)
\(\begin{aligned} \Rightarrow \quad(x+1)^{2} &=A\left(x^{2}+1\right)+(B x+c)(x) \\ x^{2}+1+2 x &=A x^{2}+A+B x^{2}+C x \\ &=(A+B) x^{2}+C x+A \end{aligned}\)
On comparing the coefficient of like powers on both sides, we get
\(A+B=1...(ii)\)
and \(\quad C=2, \quad A=1\)
From Eq. (ii), \(B=0\)
Then, \(\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C\)
\(=\sin ^{-1}(1)+\tan ^{-1}(0)+\sec ^{-1}(2)\)
\(=\sin ^{-1} \sin \frac{\pi}{2}+\tan ^{-1} \tan 0+\sec ^{-1} \sec \frac{\pi}{3}\)
\(=\frac{\pi}{2}+0+\frac{\pi}{3}=\frac{5 \pi}{6}\)
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