What is the integral of ((x))/((7x^2+3)^5) ?

The integral of ((x))/((7x^2+3)^5) is -1/(56(7x^2+3)^4)+C

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integral of ((x))/((7x^2+3)^5)

\int \frac{( x )}{( 7 x ^{2} + 3 ) ^{5}} d x

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

Solution steps

\int \frac{x}{( 7 x ^{2} + 3 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{14 u ^{5}} d u

= \int \frac{1}{14 u ^{5}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{14} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= \frac{1}{14} (- \frac{1}{4 u ^{4}})

Substitute back

u = 7 x^{2} + 3

= \frac{1}{14} (- \frac{1}{4 ( 7 x ^{2} + 3 ) ^{4}})

Simplify

\frac{1}{14} (- \frac{1}{4 ( 7 x ^{2} + 3 ) ^{4}}) : - \frac{1}{56 ( 7 x ^{2} + 3 ) ^{4}}

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

Add a constant to the solution

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

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What is the integral of ((x))/((7x^2+3)^5) ?
The integral of ((x))/((7x^2+3)^5) is -1/(56(7x^2+3)^4)+C