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

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

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integral of 1/(x^2+5)

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{5 ( 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}{5} \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}{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

x \to 2 lim (\frac{- 2}{x ( x - 2 )})

f (x) = \frac{e ^{x}}{4 x + 1}

\int \frac{3 x ^{2} + 2}{( x ^{3} + 2 x ) ^{2}} d x

y = \frac{2 x}{x + 1}, (1, 1)

\frac{d}{d x} (\frac{5 x}{x + 1})

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