What is the integral of ((4x^3+6x))/((x^4+3x^2+1)^8) ?

The integral of ((4x^3+6x))/((x^4+3x^2+1)^8) is -1/(7(x^4+3x^2+1)^7)+C

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integral of ((4x^3+6x))/((x^4+3x^2+1)^8)

\int \frac{( 4 x ^{3} + 6 x )}{( x ^{4} + 3 x ^{2} + 1 ) ^{8}} d x

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

Solution steps

\int \frac{4 x ^{3} + 6 x}{( x ^{4} + 3 x ^{2} + 1 ) ^{8}} d x

Apply u-substitution

: \int \frac{1}{u ^{8}} d u

= \int \frac{1}{u ^{8}} d u

Apply the Power Rule

: - \frac{1}{7 u ^{7}}

= - \frac{1}{7 u ^{7}}

Substitute back

u = x^{4} + 3 x^{2} + 1

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

Add a constant to the solution

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

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What is the integral of ((4x^3+6x))/((x^4+3x^2+1)^8) ?
The integral of ((4x^3+6x))/((x^4+3x^2+1)^8) is -1/(7(x^4+3x^2+1)^7)+C