What is the integral of 1/(64x^2+9) ?

The integral of 1/(64x^2+9) is 1/24 arctan(8/3 x)+C

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integral of 1/(64x^2+9)

\int \frac{1}{64 x ^{2} + 9} d x

\frac{1}{24} arctan (\frac{8}{3} x) + C

Solution steps

\int \frac{1}{64 x ^{2} + 9} d x

Apply Integral Substitution

= \int \frac{1}{24 ( 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}{24} \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}{24} arctan (u)

Substitute back

u = \frac{8}{3} x

= \frac{1}{24} arctan (\frac{8}{3} x)

Add a constant to the solution

= \frac{1}{24} arctan (\frac{8}{3} x) + C

\int csc (a x) d x

\int (1 + 4 x)^{\frac{1}{2}} d x

\int 1 t^{3} t + \frac{1}{t t} d t

What is the integral of 1/(64x^2+9) ?
The integral of 1/(64x^2+9) is 1/24 arctan(8/3 x)+C