What is the integral of (-10)/(9+4x^2) ?

The integral of (-10)/(9+4x^2) is -5/3 arctan(2/3 x)+C

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integral of (-10)/(9+4x^2)

\int \frac{- 10}{9 + 4 x ^{2}} d x

- \frac{5}{3} arctan (\frac{2}{3} x) + C

Solution steps

\int \frac{- 10}{9 + 4 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 10 \cdot \int \frac{1}{9 + 4 x ^{2}} d x

Apply Integral Substitution

= - 10 \cdot \int \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 10 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= - 10 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{2}{3} x

= - 10 \cdot \frac{1}{6} arctan (\frac{2}{3} x)

Simplify

- 10 \cdot \frac{1}{6} arctan (\frac{2}{3} x) : - \frac{5}{3} arctan (\frac{2}{3} x)

= - \frac{5}{3} arctan (\frac{2}{3} x)

Add a constant to the solution

= - \frac{5}{3} arctan (\frac{2}{3} x) + C

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

\int sec^{2} (\frac{x}{6}) d x

\int sin^{4} (\frac{x}{9}) cos (\frac{x}{9}) d x

\frac{1}{( x ^{2} + 1 )}

\int \frac{4}{1 - 4 x ^{2}} d x

What is the integral of (-10)/(9+4x^2) ?
The integral of (-10)/(9+4x^2) is -5/3 arctan(2/3 x)+C