What is the integral of 1/(36x^2+100) ?

The integral of 1/(36x^2+100) is 1/60 arctan((3x)/5)+C

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integral of 1/(36x^2+100)

\int \frac{1}{36 x ^{2} + 100} d x

\frac{1}{60} arctan (\frac{3 x}{5}) + C

Solution steps

\int \frac{1}{36 x ^{2} + 100} d x

Apply Integral Substitution

= \int \frac{1}{60 ( 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}{60} \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}{60} arctan (u)

Substitute back

u = \frac{6}{10} x

= \frac{1}{60} arctan (\frac{6}{10} x)

Simplify

\frac{1}{60} arctan (\frac{6}{10} x) : \frac{1}{60} arctan (\frac{3 x}{5})

= \frac{1}{60} arctan (\frac{3 x}{5})

Add a constant to the solution

= \frac{1}{60} arctan (\frac{3 x}{5}) + C

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What is the integral of 1/(36x^2+100) ?
The integral of 1/(36x^2+100) is 1/60 arctan((3x)/5)+C