What is the integral of 7/(9+4x^2) ?

The integral of 7/(9+4x^2) is 7/6 arctan(2/3 x)+C

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integral of 7/(9+4x^2)

\int \frac{7}{9 + 4 x ^{2}} d x

\frac{7}{6} arctan (\frac{2}{3} x) + C

Solution steps

\int \frac{7}{9 + 4 x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{9 + 4 x ^{2}} d x

Apply Integral Substitution

= 7 \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

= 7 \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)

= 7 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{2}{3} x

= 7 \cdot \frac{1}{6} arctan (\frac{2}{3} x)

Simplify

7 \cdot \frac{1}{6} arctan (\frac{2}{3} x) : \frac{7}{6} arctan (\frac{2}{3} x)

= \frac{7}{6} arctan (\frac{2}{3} x)

Add a constant to the solution

= \frac{7}{6} arctan (\frac{2}{3} x) + C

\frac{d}{d x} (ln (\frac{x}{7}))

y^{^{''}} + 5 y^{^{'}} + 6 = 1

3 e^{- 4 t}

(4 x + 3) (2 x^{2} + 7 x - 1)

\int x^{2} - 21 d x

What is the integral of 7/(9+4x^2) ?
The integral of 7/(9+4x^2) is 7/6 arctan(2/3 x)+C