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

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

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

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

\frac{1}{2} arctan (2 x) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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)

= \frac{1}{2} arctan (u)

Substitute back

u = 2 x

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

Add a constant to the solution

= \frac{1}{2} arctan (2 x) + C

\int \frac{1}{y + y} d y

y^{^{'}} = sin (x^{2})

n = 0 \sum \infty \frac{n + 4}{3 n + 2}

\int (z - 3)^{n} d z

\int_{- 1}^{1} (7 x^{3} - 2 x^{2} + 5 x - 4) d x

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