If In - - dx-x2 - a2n then In - 1 - 1 - 2n-2n- 1-a2 In is equal to -

The correct option is C 1/2na^2. x/(x^2 + a^2)^n I_n = x/(x^2 + a^2) + 2n ∫x^2/(x^2 + a^2)^n + 1dx = x/(x^2 + a^2)^n + 2n [∫ ( x^ ...

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Chain Rule of Differentiation

If In = ∫ d...

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{x}{(x^{2} + a^{2})^{n}}

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

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Solution

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Similar questions

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

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}

a_{0} + a_{3} + a_{6} + . . . =

{(x + a)}^{n}

p^{2} + q^{2}

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