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

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

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

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

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

Solution steps

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

\int \frac{( x ^{3} + 4 )}{x ^{2} + 4} d x

\int 2 e^{4 x} d x

\frac{5}{3 ( s + 3 ) ^{2} + 27}

x \to - 1 lim (\frac{x}{x - 1})

\frac{d}{d y} (\frac{x - y}{x + y})

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