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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 4 u - 2} d u

= \int \frac{1}{2 4 u - 2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{4 u - 2} d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{4 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot 2 v^{\frac{1}{2}}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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