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

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

integral of 1/(5+x^2)

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{5 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{5} arctan (u)

Substitute back

u = \frac{x}{5}

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

Add a constant to the solution

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

\int \frac{ln ( y )}{lo g _{10} ( y )} d y

f (x) = x^{4} - 17 x^{2} + 16

\int π (\frac{sin ( x )}{x})^{2} d x

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

y = ∣ x - 6 ∣, y = \frac{x}{2}

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