What is the integral of 1/(2x^2+5) ?

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

integral of 1/(2x^2+5)

\int \frac{1}{2 x ^{2} + 5} d x

\frac{1}{10} arctan (\frac{2}{5} x) + C

Solution steps

\int \frac{1}{2 x ^{2} + 5} d x

Apply Integral Substitution

= \int \frac{1}{10 ( 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}{10} \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}{10} arctan (u)

Substitute back

u = \frac{2}{5} x

= \frac{1}{10} arctan (\frac{2}{5} x)

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{10} arctan (\frac{2}{5} x)

Add a constant to the solution

= \frac{1}{10} arctan (\frac{2}{5} x) + C

\int tan (4 x) sec^{2} (4 x) d x

\int 15 x \cdot 1 - x d x

d er i v a t i v e \frac{4 - sec ( x )}{tan ( x )}

x \to 0 lim (\frac{1}{ln ( 1 + x )} - \frac{1}{x})

d er i v a t i v e (x^{4} + 3 x)^{- 1}

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