What is the integral of x/(sqrt(x+8)) ?

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

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integral of x/(sqrt(x+8))

\int \frac{x}{x + 8} d x

2 (\frac{1}{3} (x + 8)^{\frac{3}{2}} - 8 x + 8) + C

Solution steps

\int \frac{x}{x + 8} d x

Apply u-substitution

: \int 2 (u^{2} - 8) d u

= \int 2 (u^{2} - 8) 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} - 8 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

= 2 (\int u^{2} d u - \int 8 d u)

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

\int 8 d u = 8 u

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

Substitute back

u = x + 8

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

Simplify

2 (\frac{( x + 8 ) ^{3}}{3} - 8 x + 8) : 2 (\frac{1}{3} (x + 8)^{\frac{3}{2}} - 8 x + 8)

= 2 (\frac{1}{3} (x + 8)^{\frac{3}{2}} - 8 x + 8)

Add a constant to the solution

= 2 (\frac{1}{3} (x + 8)^{\frac{3}{2}} - 8 x + 8) + C

x \to 2 lim (\frac{1}{2 - 5 x})

\frac{d}{d t} (e^{4 t})

0.125 y^{^{''}} + 4 y^{^{'}} + 28 y = 0

\frac{d}{d x} (6 x^{3} cos (x))

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