For x geq 0, int sqrt{x^2+2x} , dx is equal t | vedclass

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

Question

(D) Let $I = \int \sqrt{x^2+2x} \, dx$. Completing the square, we have $x^2+2x = (x+1)^2 - 1$. So, $I = \int \sqrt{(x+1)^2 - 1} \, dx$. Using the stan

Vedclass · Accepted Answer

$\frac{x+1}{2} \sqrt{x^2+2x} - \frac{1}{2} \cosh^{-1}(x+1) + C$

Vedclass · Answer

(D) Let $I = \int \sqrt{x^2+2x} \, dx$. Completing the square, we have $x^2+2x = (x+1)^2 - 1$. So, $I = \int \sqrt{(x+1)^2 - 1} \, dx$. Using the standard integral formula $\int \sqrt{u^2-a^2} \, du = \frac{u}{2} \sqrt{u^2-a^2} - \frac{a^2}{2} \cosh^{-1} \left( \frac{u}{a} \right) + C$, where $u = x+1$ and $a = 1$: $I = \frac{x+1}{2} \sqrt{(x+1)^2-1} - \frac{1^2}{2} \cosh^{-1} \left( \frac{x+1}{1} \right) + C$. Thus, $I = \frac{x+1}{2} \sqrt{x^2+2x} - \frac{1}{2} \cosh^{-1}(x+1) + C$.

For $x \geq 0$, $\int \sqrt{x^2+2x} \, dx$ is equal to

Similar Questions

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$ then $\frac{m-l}{r}=$

$\int \frac{dx}{(1+x) \sqrt{8+7x-x^2}} = $

If $\int \sqrt{\frac{x - 5}{x - 7}} dx = A \sqrt{x^2 - 12 x + 35} + \log |x - 6 + \sqrt{x^2 - 12 x + 35}| + C$,then $A = . . . . . .$

For $n \geq 2$,if $I_n = \int \sec^n x \, dx$,then $I_4 - \frac{2}{3} I_2 =$

If $f(x) = \int \left[ \tan^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{\sin^2 2x} \right] dx$ and $f\left(\frac{\pi}{4}\right) = 0$, then $3 \left[ f\left(\frac{\pi}{6}\right) + 2 \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module