What is the integral of (2x^3+3x)/((x^4+3x^2+7)^4) ?

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

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integral of (2x^3+3x)/((x^4+3x^2+7)^4)

\int \frac{2 x ^{3} + 3 x}{( x ^{4} + 3 x ^{2} + 7 ) ^{4}} d x

- \frac{1}{6 ( x ^{4} + 3 x ^{2} + 7 ) ^{3}} + C

Solution steps

\int \frac{2 x ^{3} + 3 x}{( x ^{4} + 3 x ^{2} + 7 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{4}} d u

= \int \frac{1}{2 u ^{4}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{2} (- \frac{1}{3 u ^{3}})

Substitute back

u = x^{4} + 3 x^{2} + 7

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

Simplify

\frac{1}{2} (- \frac{1}{3 ( x ^{4} + 3 x ^{2} + 7 ) ^{3}}) : - \frac{1}{6 ( x ^{4} + 3 x ^{2} + 7 ) ^{3}}

= - \frac{1}{6 ( x ^{4} + 3 x ^{2} + 7 ) ^{3}}

Add a constant to the solution

= - \frac{1}{6 ( x ^{4} + 3 x ^{2} + 7 ) ^{3}} + C

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