What is the integral of (10x)/(sqrt(4x^2-4)) ?

The integral of (10x)/(sqrt(4x^2-4)) is 5sqrt(x^2-1)+C

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integral of (10x)/(sqrt(4x^2-4))

\int \frac{10 x}{4 x ^{2} - 4} d x

5 x^{2} - 1 + C

Solution steps

\int \frac{10 x}{4 x ^{2} - 4} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{4 x ^{2} - 4} d x

Apply u-substitution

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

= 10 \cdot \int \frac{1}{4 u - 1} d u

Take the constant out:

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

= 10 \cdot \frac{1}{4} \cdot \int \frac{1}{u - 1} d u

Apply u-substitution

: \int 2 d v

= 10 \cdot \frac{1}{4} \cdot \int 2 d v

Integral of a constant:

\int a d x = a x

= 10 \cdot \frac{1}{4} \cdot 2 v

Substitute back

= 10 \cdot \frac{1}{4} \cdot 2 x^{2} - 1

Simplify

10 \cdot \frac{1}{4} \cdot 2 x^{2} - 1 : 5 x^{2} - 1

= 5 x^{2} - 1

Add a constant to the solution

= 5 x^{2} - 1 + C

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What is the integral of (10x)/(sqrt(4x^2-4)) ?
The integral of (10x)/(sqrt(4x^2-4)) is 5sqrt(x^2-1)+C