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

Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$.

Evaluate the following integral:
$\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$

Evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$.

Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{x}{(1+x^4)^{1/4}}$ and $g(x)=f(f(f(f(x))))$. Then find the value of $18 \int_0^{\sqrt{2\sqrt{5}}} x^3 g(x) dx$.

$\int_0^{\pi / 4} \frac{\sec x}{3 \cos x+4 \sin x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module