What is the antiderivative of $\frac{x}{{(7 x^{2} + 3)}^{5}}$ $x/(7x^2+3)^5$ ?

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What is the antiderivative of $\frac{x}{{(7 x^{2} + 3)}^{5}}$ $x/(7x^2+3)^5$ ?

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Answer 1 · 2016-05-29T02:11:28

$- \frac{1}{56 {(7 x^{2} + 3)}^{4}} + C$ $-1/(56(7x^2+3)^4)+C$

Explanation:

We wish to find:

$\int \frac{x}{{(7 x^{2} + 3)}^{5}} d x$ $intx/(7x^2+3)^5dx$

This is a prime case for substitution. Let $u = 7 x^{2} + 3$ $u=7x^2+3$ so that $d u = 14 x d x$ $du=14xdx$ .

Note that we already have $x d x$ $xdx$ in the numerator, so we need to multiply the integrand by $14$ $14$ to obtain $d u$ $du$ . Balance this by multiplying the exterior of the integral by $\frac{1}{14}$ $1/14$ .

$= \frac{1}{14} \int \frac{14 x}{{(7 x^{2} + 3)}^{5}} d x$ $=1/14int(14x)/(7x^2+3)^5dx$

Now substitute in $u$ $u$ and $d u$ $du$ .

$= \frac{1}{14} \int \frac{d u}{u^{5}} = \frac{1}{14} \int u^{- 5} d u$ $=1/14int(du)/u^5=1/14intu^-5du$

Integrating $u^{- 5}$ $u^-5$ results in $\frac{u^{- 5 + 1}}{- 5 + 1} + C$ $u^(-5+1)/(-5+1)+C$ :

$= \frac{1}{14} (\frac{u^{- 4}}{- 4}) + C = - \frac{1}{56 u^{4}} + C = - \frac{1}{56 {(7 x^{2} + 3)}^{4}} + C$ $=1/14(u^-4/(-4))+C=-1/(56u^4)+C=-1/(56(7x^2+3)^4)+C$

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What is the antiderivative of $\frac{x}{{(7 x^{2} + 3)}^{5}}$ $x/(7x^2+3)^5$ ?

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Explanation:

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What is the antiderivative of $\frac{x}{{(7 x^{2} + 3)}^{5}}$ $x/(7x^2+3)^5$ ?