Evaluate the integral int frac{d x}{(x+100) s | vedclass

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

Question

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.We can rewrite the denominator as $(x+99) + 1 = (\sqrt{x+99})^2 + 1$.So,$I = \int \frac{d x}{((\sqr

Vedclass · Accepted Answer

$2 	an^{-1}(\sqrt{x+99})$

Vedclass · Answer

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.We can rewrite the denominator as $(x+99) + 1 = (\sqrt{x+99})^2 + 1$.So,$I = \int \frac{d x}{((\sqrt{x+99})^2 + 1) \sqrt{x+99}}$.Let $t = \sqrt{x+99}$. Then $t^2 = x+99$,which implies $2t \, dt = dx$.Substituting these into the integral:$I = \int \frac{2t \, dt}{(t^2 + 1)t} = \int \frac{2 \, dt}{t^2 + 1}$.Using the standard integral formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,we get:$I = 2 	an^{-1}(t) + c$.Substituting back $t = \sqrt{x+99}$:$I = 2 	an^{-1}(\sqrt{x+99}) + c$.Comparing this with $f(x) + c$,we get $f(x) = 2 	an^{-1}(\sqrt{x+99})$.

Evaluate the integral $\int \frac{d x}{(x+100) \sqrt{x+99}} = f(x) + c$. Find $f(x)$.

Similar Questions

$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to

Integrate the function $\frac{1}{x-\sqrt{x}}$.

$\int \sec x \log(\sec x + \tan x) \, dx = $

Evaluate the integral $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = -f(x) + c$. Find $f(x)$.

If $\int \frac{x^{49} \tan ^{-1}(x^{50})}{1+x^{100}} d x=k(\tan ^{-1}(x^{50}))^2+c$, then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module