What is the integral of (8x)/(sqrt(x^2+16)) ?

The integral of (8x)/(sqrt(x^2+16)) is 8sqrt(x^2+16)+C

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integral of (8x)/(sqrt(x^2+16))

\int \frac{8 x}{x ^{2} + 16} d x

8 x^{2} + 16 + C

Solution steps

\int \frac{8 x}{x ^{2} + 16} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{x ^{2} + 16} d x

Apply u-substitution

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

= 8 \cdot \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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} + 16

= 8 \cdot \frac{1}{2} \cdot 2 (x^{2} + 16)^{\frac{1}{2}}

Simplify

8 \cdot \frac{1}{2} \cdot 2 (x^{2} + 16)^{\frac{1}{2}} : 8 x^{2} + 16

= 8 x^{2} + 16

Add a constant to the solution

= 8 x^{2} + 16 + C

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What is the integral of (8x)/(sqrt(x^2+16)) ?
The integral of (8x)/(sqrt(x^2+16)) is 8sqrt(x^2+16)+C