-x-a2-x2-n-a2-x2 dx n - 0-

Put x+√(a^2+x^2)=t then [ 1+2x/2√(a^2+x^2) ] dx = dt ∫[x+√(a^2+x^2)]^n/√(a^2+x^2) dx = ∫ t^n 1 dt = t^n/n+c = 1/n(x+√(a^2+x^2))^n+c ...

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Integration by Substitution

∫[x+√a2+x2]n/...

Question

$\int \frac{[x + \sqrt{a^{2} + x^{2}}]^{n}}{\sqrt{a^{2} + x^{2}}} d x (n \neq 0) =$

\frac{1}{n + 1} (x + \sqrt{a^{2} + x^{2}})^{n + 1} + c

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\frac{1}{n . (x + \sqrt{a^{2} + x^{2}})^{n}} + c

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(x + \sqrt{a^{2} + x^{2}})^{n} + c

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\frac{1}{n} (x + \sqrt{a^{2} + x^{2}})^{n} + c

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Solution

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\int \frac{[x + \sqrt{a^{2} + x^{2}}]^{n}}{\sqrt{a^{2} + x^{2}}} d x (n \neq 0) =

y = [x + \sqrt{x^{2} + a^{2}}]^{n}

\frac{d y}{d x} = \frac{n y}{\sqrt{x^{2} + a^{2}}}

\int \frac{[x + \sqrt{a^{2} + x^{2}}]^{n}}{\sqrt{a^{2} + x^{2}}} d x (n \neq 0) =

If $y = {[x + \sqrt{x^{2} + a^{2}}]}^{n}, t h e n \frac{d y}{d x}$ is equal to

\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = l n | x + \sqrt{x^{2} + a^{2}} | + C

\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = l n | x - \sqrt{x^{2} - a^{2}} | + C

\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} l n ∣ ∣ \frac{x - a}{x + a} ∣ ∣ + C

\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l n ∣ ∣ \frac{x + a}{x - a} ∣ ∣ + C

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