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

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

Popular >

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

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

\frac{1}{4} arctan (\frac{x}{2}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (u)

Substitute back

u = \frac{2}{2 2} x

= \frac{1}{4} arctan (\frac{2}{2 2} x)

Simplify

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

= \frac{1}{4} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x}{2}) + C

\int 11 e^{10 x} d x

\frac{d y}{d x} = \frac{( y ^{2} - 2 x y )}{( x ^{2} - 2 x y )}

\int \frac{x ^{2} - 12 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

(2 t - 1)^{3}

\int - 7 x e^{x^{2}} d x

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