What is the integral of x/(sqrt(4x^2-9)) ?

The integral of x/(sqrt(4x^2-9)) is 1/4 sqrt(4x^2-9)+C

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integral of x/(sqrt(4x^2-9))

\int \frac{x}{4 x ^{2} - 9} d x

\frac{1}{4} 4 x^{2} - 9 + C

Solution steps

\int \frac{x}{4 x ^{2} - 9} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 4 x^{2} - 9

= \frac{1}{8} \cdot 2 (4 x^{2} - 9)^{\frac{1}{2}}

Simplify

\frac{1}{8} \cdot 2 (4 x^{2} - 9)^{\frac{1}{2}} : \frac{1}{4} 4 x^{2} - 9

= \frac{1}{4} 4 x^{2} - 9

Add a constant to the solution

= \frac{1}{4} 4 x^{2} - 9 + C

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What is the integral of x/(sqrt(4x^2-9)) ?
The integral of x/(sqrt(4x^2-9)) is 1/4 sqrt(4x^2-9)+C