What is the integral of 8/(36x^2+4) ?

The integral of 8/(36x^2+4) is 2/3 arctan(3x)+C

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integral of 8/(36x^2+4)

\int \frac{8}{36 x ^{2} + 4} d x

\frac{2}{3} arctan (3 x) + C

Solution steps

\int \frac{8}{36 x ^{2} + 4} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{36 x ^{2} + 4} d x

Apply Integral Substitution

= 8 \cdot \int \frac{1}{3 ( 4 u ^{2} + 4 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{4 u ^{2} + 4} d u

\frac{1}{4 u ^{2} + 4} = \frac{1}{4} \frac{1}{u ^{2} + 1}

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{4} \cdot \frac{1}{u ^{2} + 1} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 8 \cdot \frac{1}{3} \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{6}{2} x

= 8 \cdot \frac{1}{3} \cdot \frac{1}{4} arctan (\frac{6}{2} x)

Simplify

8 \cdot \frac{1}{3} \cdot \frac{1}{4} arctan (\frac{6}{2} x) : \frac{2}{3} arctan (3 x)

= \frac{2}{3} arctan (3 x)

Add a constant to the solution

= \frac{2}{3} arctan (3 x) + C

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What is the integral of 8/(36x^2+4) ?
The integral of 8/(36x^2+4) is 2/3 arctan(3x)+C