What is the integral of 1/((x^2+25)) ?

The integral of 1/((x^2+25)) is 1/5 arctan(x/5)+C

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integral of 1/((x^2+25))

\int \frac{1}{( x ^{2} + 25 )} d x

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

Solution steps

\int \frac{1}{x ^{2} + 25} 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

y = \frac{5}{7} x + \frac{4}{3}

y^{^{''}} + 2 y = 3, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d}{d x} (lo g_{5} (2 x^{3} + x^{2}))

\int \frac{13}{r ^{2} + 1} d r

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 8 x - y}

What is the integral of 1/((x^2+25)) ?
The integral of 1/((x^2+25)) is 1/5 arctan(x/5)+C