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

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(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

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