What is the integral of x/(sqrt(x^2+8)) ?

The integral of x/(sqrt(x^2+8)) is sqrt(x^2+8)+C

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integral of x/(sqrt(x^2+8))

\int \frac{x}{x ^{2} + 8} d x

x^{2} + 8 + C

Solution steps

\int \frac{x}{x ^{2} + 8} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = x^{2} + 8

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

Simplify

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

= x^{2} + 8

Add a constant to the solution

= x^{2} + 8 + C

\frac{d}{d x} (x^{8} arctan (x^{3}))

y^{^{'}} = k (y - 70)

\int \frac{ln ( 8 x )}{x} d x

\frac{\partial}{\partial z} (e^{yz})

y = ln (\frac{x - 1}{x + 3}) - x

What is the integral of x/(sqrt(x^2+8)) ?
The integral of x/(sqrt(x^2+8)) is sqrt(x^2+8)+C