What is the integral of 5xsqrt(x+4) ?

The integral of 5xsqrt(x+4) is 2(x+4)^{5/2}-40/3 (x+4)^{3/2}+C

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integral of 5xsqrt(x+4)

\int 5 x x + 4 d x

2 (x + 4)^{\frac{5}{2}} - \frac{40}{3} (x + 4)^{\frac{3}{2}} + C

Solution steps

\int 5 x x + 4 d x

Take the constant out:

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

= 5 \cdot \int x x + 4 d x

Apply u-substitution

: \int (u - 4) u d u

= 5 \cdot \int (u - 4) u d u

Expand

(u - 4) u : u^{\frac{3}{2}} - 4 u

= 5 \cdot \int u^{\frac{3}{2}} - 4 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= 5 (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = x + 4

= 5 (\frac{2}{5} (x + 4)^{\frac{5}{2}} - \frac{8}{3} (x + 4)^{\frac{3}{2}})

Simplify

5 (\frac{2}{5} (x + 4)^{\frac{5}{2}} - \frac{8}{3} (x + 4)^{\frac{3}{2}}) : 2 (x + 4)^{\frac{5}{2}} - \frac{40}{3} (x + 4)^{\frac{3}{2}}

= 2 (x + 4)^{\frac{5}{2}} - \frac{40}{3} (x + 4)^{\frac{3}{2}}

Add a constant to the solution

= 2 (x + 4)^{\frac{5}{2}} - \frac{40}{3} (x + 4)^{\frac{3}{2}} + C

\frac{d}{d x} ((1 - x)^{\frac{2}{3}})

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What is the integral of 5xsqrt(x+4) ?
The integral of 5xsqrt(x+4) is 2(x+4)^{5/2}-40/3 (x+4)^{3/2}+C