What is the integral of (7x^6+4x^3)/((x^7+x^4)^3) ?

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

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integral of (7x^6+4x^3)/((x^7+x^4)^3)

\int \frac{7 x ^{6} + 4 x ^{3}}{( x ^{7} + x ^{4} ) ^{3}} d x

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

Solution steps

\int \frac{7 x ^{6} + 4 x ^{3}}{( x ^{7} + x ^{4} ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{2 u ^{2}}

Substitute back

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

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

Add a constant to the solution

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

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