What is the integral of sqrt(x)(x^{3/2}-10)^4 ?

The integral of sqrt(x)(x^{3/2}-10)^4 is 2/15 (x^{3/2}-10)^5+C

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integral of sqrt(x)(x^{3/2}-10)^4

\int x (x^{\frac{3}{2}} - 10)^{4} d x

\frac{2}{15} (x^{\frac{3}{2}} - 10)^{5} + C

Solution steps

\int x (x^{\frac{3}{2}} - 10)^{4} d x

Apply u-substitution

: \int 2 u^{2} (u^{3} - 10)^{4} d u

= \int 2 u^{2} (u^{3} - 10)^{4} d u

Take the constant out:

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

= 2 \cdot \int u^{2} (u^{3} - 10)^{4} d u

Apply u-substitution

: \int \frac{v ^{4}}{3} d v

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{v ^{5}}{5}

= 2 \cdot \frac{1}{3} \cdot \frac{v ^{5}}{5}

Substitute back

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

Simplify

2 \cdot \frac{1}{3} \cdot \frac{( ( x ) ^{3} - 10 ) ^{5}}{5} : \frac{2}{15} (x^{\frac{3}{2}} - 10)^{5}

= \frac{2}{15} (x^{\frac{3}{2}} - 10)^{5}

Add a constant to the solution

= \frac{2}{15} (x^{\frac{3}{2}} - 10)^{5} + C

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What is the integral of sqrt(x)(x^{3/2}-10)^4 ?
The integral of sqrt(x)(x^{3/2}-10)^4 is 2/15 (x^{3/2}-10)^5+C