What is the integral of 1/(9+16x^2) ?

The integral of 1/(9+16x^2) is 1/12 arctan(4/3 x)+C

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integral of 1/(9+16x^2)

\int \frac{1}{9 + 16 x ^{2}} d x

\frac{1}{12} arctan (\frac{4}{3} x) + C

Solution steps

\int \frac{1}{9 + 16 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{12 ( 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}{12} \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}{12} arctan (u)

Substitute back

u = \frac{4}{3} x

= \frac{1}{12} arctan (\frac{4}{3} x)

Add a constant to the solution

= \frac{1}{12} arctan (\frac{4}{3} x) + C

\frac{d}{d x} (x ln (sin (x)))

\int_{0}^{\infty} e^{- x} d x

\int x cos (x^{2}) d x

f (x) = 9 + \frac{3}{x}

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

What is the integral of 1/(9+16x^2) ?
The integral of 1/(9+16x^2) is 1/12 arctan(4/3 x)+C