What is the integral of x/(sqrt(81x^2+64)) ?

The integral of x/(sqrt(81x^2+64)) is 1/81 sqrt(81x^2+64)+C

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integral of x/(sqrt(81x^2+64))

\int \frac{x}{81 x ^{2} + 64} d x

\frac{1}{81} 81 x^{2} + 64 + C

Solution steps

\int \frac{x}{81 x ^{2} + 64} d x

Apply u-substitution

: \int \frac{1}{162 u} d u

= \int \frac{1}{162 u} d u

Take the constant out:

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

= \frac{1}{162} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{162} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 81 x^{2} + 64

= \frac{1}{162} \cdot 2 (81 x^{2} + 64)^{\frac{1}{2}}

Simplify

\frac{1}{162} \cdot 2 (81 x^{2} + 64)^{\frac{1}{2}} : \frac{1}{81} 81 x^{2} + 64

= \frac{1}{81} 81 x^{2} + 64

Add a constant to the solution

= \frac{1}{81} 81 x^{2} + 64 + C

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What is the integral of x/(sqrt(81x^2+64)) ?
The integral of x/(sqrt(81x^2+64)) is 1/81 sqrt(81x^2+64)+C