What is the integral of (5x^3)/((5-3x^4)^8) ?

The integral of (5x^3)/((5-3x^4)^8) is 5/(84(5-3x^4)^7)+C

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integral of (5x^3)/((5-3x^4)^8)

\int \frac{5 x ^{3}}{( 5 - 3 x ^{4} ) ^{8}} d x

\frac{5}{84 ( 5 - 3 x ^{4} ) ^{7}} + C

Solution steps

\int \frac{5 x ^{3}}{( 5 - 3 x ^{4} ) ^{8}} 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 ^{3}}{( 5 - 3 x ^{4} ) ^{8}} d x

Apply u-substitution

: \int - \frac{1}{12 u ^{8}} d u

= 5 \cdot \int - \frac{1}{12 u ^{8}} d u

Take the constant out:

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

= 5 (- \frac{1}{12} \cdot \int \frac{1}{u ^{8}} d u)

Apply the Power Rule

: - \frac{1}{7 u ^{7}}

= 5 (- \frac{1}{12} (- \frac{1}{7 u ^{7}}))

Substitute back

u = 5 - 3 x^{4}

= 5 (- \frac{1}{12} (- \frac{1}{7 ( 5 - 3 x ^{4} ) ^{7}}))

Simplify

5 (- \frac{1}{12} (- \frac{1}{7 ( 5 - 3 x ^{4} ) ^{7}})) : \frac{5}{84 ( 5 - 3 x ^{4} ) ^{7}}

= \frac{5}{84 ( 5 - 3 x ^{4} ) ^{7}}

Add a constant to the solution

= \frac{5}{84 ( 5 - 3 x ^{4} ) ^{7}} + C

\frac{\partial}{\partial x} (sin (x^{2} + y^{2}))

x \to 2 + lim (3 + ∣ 2 x - 4 ∣)

\frac{d}{d y} (cos (x) - y)

(cot (y) + x^{2}) d x = x csc^{2} (y) d y

\frac{1}{x - 2}

What is the integral of (5x^3)/((5-3x^4)^8) ?
The integral of (5x^3)/((5-3x^4)^8) is 5/(84(5-3x^4)^7)+C