What is the integral of 4x(x^2+25000)^{-2/3} ?

The integral of 4x(x^2+25000)^{-2/3} is 6(x^2+25000)^{1/3}+C

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integral of 4x(x^2+25000)^{-2/3}

\int 4 x (x^{2} + 25000)^{- \frac{2}{3}} d x

6 (x^{2} + 25000)^{\frac{1}{3}} + C

Solution steps

\int 4 x (x^{2} + 25000)^{- \frac{2}{3}} d x

Take the constant out:

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

= 4 \cdot \int x (x^{2} + 25000)^{- \frac{2}{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{2}{3}}} d u

= 4 \cdot \int \frac{1}{2 u ^{\frac{2}{3}}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{2} \cdot 3 u^{\frac{1}{3}}

Substitute back

u = x^{2} + 25000

= 4 \cdot \frac{1}{2} \cdot 3 (x^{2} + 25000)^{\frac{1}{3}}

Simplify

4 \cdot \frac{1}{2} \cdot 3 (x^{2} + 25000)^{\frac{1}{3}} : 6 (x^{2} + 25000)^{\frac{1}{3}}

= 6 (x^{2} + 25000)^{\frac{1}{3}}

Add a constant to the solution

= 6 (x^{2} + 25000)^{\frac{1}{3}} + C

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What is the integral of 4x(x^2+25000)^{-2/3} ?
The integral of 4x(x^2+25000)^{-2/3} is 6(x^2+25000)^{1/3}+C