What is the integral of (3x^4)/((2x^5-5)^4) ?

The integral of (3x^4)/((2x^5-5)^4) is -1/(10(-5+2x^5)^3)+C

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integral of (3x^4)/((2x^5-5)^4)

\int \frac{3 x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

- \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}} + C

Solution steps

\int \frac{3 x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{10 u ^{4}} d u

= 3 \cdot \int \frac{1}{10 u ^{4}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{10} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{10} (- \frac{1}{3 u ^{3}})

Substitute back

u = - 5 + 2 x^{5}

= 3 \cdot \frac{1}{10} (- \frac{1}{3 ( - 5 + 2 x ^{5} ) ^{3}})

Simplify

3 \cdot \frac{1}{10} (- \frac{1}{3 ( - 5 + 2 x ^{5} ) ^{3}}) : - \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}}

= - \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}}

Add a constant to the solution

= - \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}} + C

\int 6 t^{2} d t

x^{3} e^{- 2 x}

\int e^{6 x - 7} d x

8 cos (t) - 9 sin (t)

P (t) = 566 + (37 t^{0.6} - 106)

What is the integral of (3x^4)/((2x^5-5)^4) ?
The integral of (3x^4)/((2x^5-5)^4) is -1/(10(-5+2x^5)^3)+C