What is the integral of 8/(4x^2+1) ?

The integral of 8/(4x^2+1) is 4arctan(2x)+C

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integral of 8/(4x^2+1)

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

4 arctan (2 x) + C

Solution steps

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

Apply Integral Substitution

= 8 \cdot \int \frac{1}{2 ( 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}{2} \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}{2} arctan (u)

Substitute back

u = 2 x

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

Simplify

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

= 4 arctan (2 x)

Add a constant to the solution

= 4 arctan (2 x) + C

y = 1 - x^{2}, y = 29 - 11 x

2 (3 x - 8)^{4} - 5 x + 6

x \to 1 + lim (cos (π x))

x \to 3 lim (\frac{x}{x ^{2} + 1})

\int (π - x) cos (n x) d x

What is the integral of 8/(4x^2+1) ?
The integral of 8/(4x^2+1) is 4arctan(2x)+C