int {sqrt {{x^2} + {a^2}} ,dx} is equal to | vedclass

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

Question

(B) Let $I = \int {\sqrt {{x^2} + {a^2}} \,dx} $. Using integration by parts,taking $1$ as the second function: $I = \sqrt {{x^2} + {a^2}} \cdot x - \

Vedclass · Accepted Answer

$\frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$

Vedclass · Answer

(B) Let $I = \int {\sqrt {{x^2} + {a^2}} \,dx} $. Using integration by parts,taking $1$ as the second function: $I = \sqrt {{x^2} + {a^2}} \cdot x - \int {\frac{x}{{\sqrt {{x^2} + {a^2}} }} \cdot x \,dx} $ $I = x\sqrt {{x^2} + {a^2}} - \int {\frac{{{x^2} + {a^2} - {a^2}}}{{\sqrt {{x^2} + {a^2}} }} \,dx} $ $I = x\sqrt {{x^2} + {a^2}} - \int {\sqrt {{x^2} + {a^2}} \,dx} + {a^2}\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} $ $I = x\sqrt {{x^2} + {a^2}} - I + {a^2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$ $2I = x\sqrt {{x^2} + {a^2}} + {a^2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$ $I = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$.

$\int {\sqrt {{x^2} + {a^2}} \,dx} $ is equal to

Similar Questions

Find the following integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$

Find the integral of the function $\frac{1}{\cos (x-a) \cos (x-b)}$.

If $\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx = g(x) + c$ and $g(1) = 0$,then $g\left(\frac{1}{2}\right)$ is equal to

If $\int \sec ^2 x \operatorname{cosec}^4 x \, dx = -\frac{1}{3} \cot ^3 x + k \tan x - 2 \cot x + C$,then $k$ is equal to

$k \in N, \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module