What is the integral of 1/(25+36x^2) ?

The integral of 1/(25+36x^2) is 1/30 arctan(6/5 x)+C

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integral of 1/(25+36x^2)

\int \frac{1}{25 + 36 x ^{2}} d x

\frac{1}{30} arctan (\frac{6}{5} x) + C

Solution steps

\int \frac{1}{25 + 36 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{30 ( 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}{30} \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}{30} arctan (u)

Substitute back

u = \frac{6}{5} x

= \frac{1}{30} arctan (\frac{6}{5} x)

Add a constant to the solution

= \frac{1}{30} arctan (\frac{6}{5} x) + C

\int \frac{2 x}{( x ^{2} + 5 ) ^{2}} d x

\int cos^{3} (z) d z

\int sec^{2} (x) e^{1 + t a n (x)} d x

\int - 1 ln (\frac{2}{5} x) d x

\int (x - 1) (x + 5) d x

What is the integral of 1/(25+36x^2) ?
The integral of 1/(25+36x^2) is 1/30 arctan(6/5 x)+C