What is the integral of (5x^4)/((10+x^5)^4) ?

The integral of (5x^4)/((10+x^5)^4) is -1/(3(10+x^5)^3)+C

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integral of (5x^4)/((10+x^5)^4)

\int \frac{5 x ^{4}}{( 10 + x ^{5} ) ^{4}} d x

- \frac{1}{3 ( 10 + x ^{5} ) ^{3}} + C

Solution steps

\int \frac{5 x ^{4}}{( 10 + x ^{5} ) ^{4}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{4}}{( 10 + x ^{5} ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{5 u ^{4}} d u

= 5 \cdot \int \frac{1}{5 u ^{4}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

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

= 5 \cdot \frac{1}{5} (- \frac{1}{3 u ^{3}})

Substitute back

u = 10 + x^{5}

= 5 \cdot \frac{1}{5} (- \frac{1}{3 ( 10 + x ^{5} ) ^{3}})

Simplify

5 \cdot \frac{1}{5} (- \frac{1}{3 ( 10 + x ^{5} ) ^{3}}) : - \frac{1}{3 ( 10 + x ^{5} ) ^{3}}

= - \frac{1}{3 ( 10 + x ^{5} ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( 10 + x ^{5} ) ^{3}} + C

\int \frac{sin ( x )}{sec ( x )} d x

\int_{0}^{2} \frac{1}{( x - 2 ) ^{2}} d x

\int - \frac{y}{( x + 2 ) ^{2}} d x

f (x) = x^{3} + 3 x, at x = 2

\frac{\partial}{\partial x} (x \cdot cos (y))

What is the integral of (5x^4)/((10+x^5)^4) ?
The integral of (5x^4)/((10+x^5)^4) is -1/(3(10+x^5)^3)+C