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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{3} + 4 x

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

Add a constant to the solution

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

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

\frac{3}{x + 1}

f (\frac{tan ( 1 )}{x})^{2}

3 x^{- 4}

\frac{1}{( x ^{2} + c )}

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