What is the integral of (8x)/(sqrt(2x^2+1)) ?

The integral of (8x)/(sqrt(2x^2+1)) is 4sqrt(2x^2+1)+C

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integral of (8x)/(sqrt(2x^2+1))

\int \frac{8 x}{2 x ^{2} + 1} d x

4 2 x^{2} + 1 + C

Solution steps

\int \frac{8 x}{2 x ^{2} + 1} 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}{2 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 8 \cdot \int \frac{1}{4 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}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

= 4 2 x^{2} + 1

Add a constant to the solution

= 4 2 x^{2} + 1 + C

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What is the integral of (8x)/(sqrt(2x^2+1)) ?
The integral of (8x)/(sqrt(2x^2+1)) is 4sqrt(2x^2+1)+C