If In- -dx- x2-a2 n then In-1-1-2n-2n-1-a2 In is equal to

The correct option is D 1/2n a^2.1/ ( x^2+a^2 )^n Using integration by parts I_n= x/ ( x^2+a^2 )^n+2n∫x^2/ ( x^2+a^2 )^n+1dx = x/ ( x^2+a^2 )^n+2 ...

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Fundamental Laws of Logarithms

If In= ∫dx/...

Question

If $I_{n} = \int \frac{d x}{{(x^{2} + a^{2})}^{n}}$ then $I_{n + 1} + \frac{1 - 2 n}{2 n} . \frac{1}{a^{2}}$ $I_{n}$ is equal to

{\frac{x}{x^{2} + a^{2}}}^{n}

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

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

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

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Solution

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Similar questions

I_{n} = \int \frac{d x}{(x^{2} + a^{2})^{n}}

I_{n + 1} + \frac{1 - 2 n}{2 n} . \frac{1}{a^{2}}

I_{n}

a_{1,} a_{2,} a_{3,} . . . . . . . . a_{2 n + 1}

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n + 1} - a_{2}}{a_{2 n + 1} + a_{2}} + . . . . . . + \frac{a_{2 n + 1} - a_{n}}{a_{2 n + 1} + a_{n}}

{(x + a)}^{n}

p^{2} + q^{2}

a_{1}, a_{2}, a_{3}, . . . . a_{2 n + 1}

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n} - a_{2}}{a_{2 n} + a_{2}} + . . . + \frac{a_{n + 2} - a_{n}}{a_{n + 2} + a_{n}}

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . . + a_{2 n} x^{2 n}

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