What is the integral of (5x^4+6)/((x^5+6x)^2) ?

The integral of (5x^4+6)/((x^5+6x)^2) is -1/(x^5+6x)+C

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integral of (5x^4+6)/((x^5+6x)^2)

\int \frac{5 x ^{4} + 6}{( x ^{5} + 6 x ) ^{2}} d x

- \frac{1}{x ^{5} + 6 x} + C

Solution steps

\int \frac{5 x ^{4} + 6}{( x ^{5} + 6 x ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x^{5} + 6 x

= - \frac{1}{x ^{5} + 6 x}

Add a constant to the solution

= - \frac{1}{x ^{5} + 6 x} + C

\frac{d y}{d x} = x^{2} + x

\frac{d}{d x} (x^{4} sin (x))

\int 2 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

\int \frac{x}{2 + 2 x} d x

x \to 0 lim ((e^{x} - 1)^{x})

What is the integral of (5x^4+6)/((x^5+6x)^2) ?
The integral of (5x^4+6)/((x^5+6x)^2) is -1/(x^5+6x)+C