Question

integrate x power 15 by 1 + x power 32​

Answer 1

Integration

$\int\ \frac{x^{15} }{1+x^{32} }.dx$

$\int\ \frac{x^{15} }{1+(x^{16} )^{2}} . dx .......................... Eq.(1)$

Let,

$u = x^{16} \\\\\frac{du}{dx} = 16x^{15\\}$

$\boxed{ \frac{du}{16} = x^{15}.dx }}$

Eq.(1) becomes :-

$\int\ \frac{1}{1+u^{2} } .\frac{du}{16}$ = $\frac{1}{16}$  $\int\ \frac{1}{1+u^{2} } .du$

$= \frac{1}{16}tan^{-1} (u) + C$

$\boxed{ = \frac{1}{16}tan^{-1} (x^{16} ) + C}}$

Answer 2

$\mathrm{\int \dfrac{x^{15}}{1+x^{32}}\:dx=\dfrac{1}{16}\:tan^{-1}(x^{16})+c}$

where $\mathrm{c}$ is constant of integration.

Step-by-step explanation:

We have to find

$\mathrm{\quad\quad\quad\quad \int \dfrac{x^{15}}{1+x^{32}}\:dx=\:?}$

Let $\mathrm{x^{16}=t}$ such that

$\mathrm{\quad\quad\quad 16\:x^{15}\:dx=dt}$

$\implies \mathrm{x^{15}\:dx=\dfrac{1}{16}\:dt}$

$\therefore \mathrm{\int \dfrac{x^{15}}{1+x^{32}}\:dx}$

$\mathrm{\quad =\int \dfrac{x^{15}\:dx}{1+(x^{16})^{2}}}$

$\mathrm{\quad =\int \dfrac{\dfrac{dt}{16}}{1+t^{2}}}$

$\mathrm{\quad =\dfrac{1}{16} \int \dfrac{dt}{1+t^{2}}}$

$\mathrm{\quad =\dfrac{1}{16}\: tan^{-1}(t)+c}$

where $\mathrm{c}$ is constant of integration.

$\mathrm{\quad =\dfrac{1}{16}\:tan^{-1}(x^{16})+c}$

This is the required integral.

Integration formula:

$\mathrm{\quad\quad\quad \int \dfrac{dx}{1+x^{2}}=tan^{-1}(x)+c}$

where $\mathrm{c}$ is constant of integration.

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