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Question: \[\int_{8}^{15}{\frac{dx}{(x - 3)\sqrt{x + 1}} =}\]...

∫815dx(x−3)x+1=\int_{8}^{15}{\frac{dx}{(x - 3)\sqrt{x + 1}} =}

A

12log⁡53\frac{1}{2}\log\frac{5}{3}

B

13log⁡53\frac{1}{3}\log\frac{5}{3}

C

12log⁡35\frac{1}{2}\log\frac{3}{5}

D

15log⁡35\frac{1}{5}\log\frac{3}{5}

Answer

12log⁡53\frac{1}{2}\log\frac{5}{3}

Explanation

Solution

I=∫815dx(x−3)x+1I = \int_{8}^{15}\frac{dx}{(x - 3)\sqrt{x + 1}}

Put x=tan⁡2θ⇒θ=tan⁡−1xx = \tan^{2}\theta \Rightarrow \theta = \tan^{- 1}\sqrt{x}

dx=2tan⁡θsec⁡2θdθdx = 2\tan\theta\sec^{2}\theta d\theta

∴I=∫tan⁡−18tan⁡−1152tan⁡θsec⁡2θ(tan⁡2θ−3)tan⁡2θ+1dθ\therefore I = \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan\theta\sec^{2}\theta}{(\tan^{2}\theta - 3)\sqrt{\tan^{2}\theta + 1}}d\theta}

=∫tan⁡−18tan⁡−1152tan⁡θsec⁡2θ(sec⁡2θ−4)sec⁡θdθ= \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan\theta\sec^{2}\theta}{(\sec^{2}\theta - 4)\sec\theta}d\theta}

=∫tan⁡−18tan⁡−1152tan⁡θsec⁡θ(sec⁡2θ−4)dθ= \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan\theta\sec\theta}{(\sec^{2}\theta - 4)}d\theta}

=∫tan⁡−18tan⁡−1152tan⁡θsec⁡θ(sec⁡θ−2)(sec⁡θ+2)dθ= \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan\theta\sec\theta}{(\sec\theta - 2)(\sec\theta + 2)}d\theta}

=[12log⁡(sec⁡θ−2)(sec⁡θ+2)]tan⁡−18tan⁡−115= \left\lbrack \frac{1}{2}\log\frac{(\sec\theta - 2)}{(\sec\theta + 2)} \right\rbrack_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}

=12[log⁡26−log⁡15]=12log⁡53= \frac{1}{2}\left\lbrack \log\frac{2}{6} - \log\frac{1}{5} \right\rbrack = \frac{1}{2}\log\frac{5}{3}.