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

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

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integral of 1/(25+x^2)

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

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

Solution steps

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

\frac{d}{d x} (\frac{x ^{2} y ^{2}}{2} - 2 x y - y^{2})

(1 + y) d y = 2 t d t

\int \frac{x ^{2}}{16 - x ^{2}} d x

3 1 + x

(x - 1)^{\frac{1}{2}}

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