int_{8}^{15} frac{dx}{(x - 3)sqrt{x + 1}} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_{8}^{15} \frac{dx}{(x - 3)\sqrt{x + 1}}$.Substitute $t = \sqrt{x + 1}$,then $t^2 = x + 1$,so $x = t^2 - 1$ and $dx = 2t \, dt$.When

Vedclass · Accepted Answer

$\frac{1}{2}\log \frac{5}{3}$

Vedclass · Answer

(A) Let $I = \int_{8}^{15} \frac{dx}{(x - 3)\sqrt{x + 1}}$.Substitute $t = \sqrt{x + 1}$,then $t^2 = x + 1$,so $x = t^2 - 1$ and $dx = 2t \, dt$.When $x = 8$,$t = \sqrt{8 + 1} = 3$. When $x = 15$,$t = \sqrt{15 + 1} = 4$.Substituting these into the integral:$I = \int_{3}^{4} \frac{2t \, dt}{(t^2 - 1 - 3)t} = \int_{3}^{4} \frac{2 \, dt}{t^2 - 4}$.Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} ight| + C$:$I = 2 	imes \frac{1}{2(2)} \left[ \log \left| \frac{t - 2}{t + 2} ight| ight]_{3}^{4}$.$I = \frac{1}{2} \left[ \log \left( \frac{4 - 2}{4 + 2} ight) - \log \left( \frac{3 - 2}{3 + 2} ight) ight]$.$I = \frac{1}{2} \left[ \log \left( \frac{2}{6} ight) - \log \left( \frac{1}{5} ight) ight] = \frac{1}{2} \left[ \log \left( \frac{1}{3} ight) - \log \left( \frac{1}{5} ight) ight]$.$I = \frac{1}{2} \log \left( \frac{1/3}{1/5} ight) = \frac{1}{2} \log \frac{5}{3}$.

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

Similar Questions

Which of the following is/are correct?

If $\int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$,then ${e^x} = $

$\int_0^1 \frac{1}{2+\sqrt{x}} \, dx =$

$\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6} \Rightarrow x=$

$\int_{1}^{2} \frac{\cos(\log x)}{x} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module