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

The integral of x(x+4)^{-3/2} is 2(x+4)^{1/2}+8/((x+4)^{1/2)}+C

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

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

2 (x + 4)^{\frac{1}{2}} + \frac{8}{( x + 4 ) ^{\frac{1}{2}}} + C

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int \frac{u - 4}{u ^{\frac{3}{2}}} d u

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

Expand

\frac{u - 4}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{4}{u ^{\frac{3}{2}}}

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

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

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{4}{u ^{\frac{3}{2}}} d u = - \frac{8}{u ^{\frac{1}{2}}}

= 2 u^{\frac{1}{2}} - (- \frac{8}{u ^{\frac{1}{2}}})

Substitute back

u = x + 4

= 2 (x + 4)^{\frac{1}{2}} - (- \frac{8}{( x + 4 ) ^{\frac{1}{2}}})

Simplify

= 2 (x + 4)^{\frac{1}{2}} + \frac{8}{( x + 4 ) ^{\frac{1}{2}}}

Add a constant to the solution

= 2 (x + 4)^{\frac{1}{2}} + \frac{8}{( x + 4 ) ^{\frac{1}{2}}} + C

n = 1 \sum \infty (- 1)^{n} n^{- 3}

\int \frac{1}{x ^{2} x ^{2} - 81} d x

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

\frac{- 8 x}{y} = 0

x \to 1 lim (x^{3} (5 - x))

What is the integral of x(x+4)^{-3/2} ?
The integral of x(x+4)^{-3/2} is 2(x+4)^{1/2}+8/((x+4)^{1/2)}+C