What is the integral of (16x)/(sqrt(8x^2+1)) ?

The integral of (16x)/(sqrt(8x^2+1)) is 2sqrt(8x^2+1)+C

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integral of (16x)/(sqrt(8x^2+1))

\int \frac{16 x}{8 x ^{2} + 1} d x

2 8 x^{2} + 1 + C

Solution steps

\int \frac{16 x}{8 x ^{2} + 1} d x

Take the constant out:

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

= 16 \cdot \int \frac{x}{8 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{16 u} d u

= 16 \cdot \int \frac{1}{16 u} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 16 \cdot \frac{1}{16} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 8 x^{2} + 1

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

Simplify

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

= 2 8 x^{2} + 1

Add a constant to the solution

= 2 8 x^{2} + 1 + C

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What is the integral of (16x)/(sqrt(8x^2+1)) ?
The integral of (16x)/(sqrt(8x^2+1)) is 2sqrt(8x^2+1)+C