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

\int \frac{x}{x ^{2} - 8} d x

q^{^{''}} + 2 q^{^{'}} + 4 q = 50 cos (t)

1 + x

y^{^{'}} = 4 x y^{3}

cos^{2} (x) sin (x) \frac{d y}{d x} + (cos^{3} (x)) y = 1

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