What is the integral of 3/(18+2x^2) ?

The integral of 3/(18+2x^2) is 1/2 arctan(1/3 x)+C

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integral of 3/(18+2x^2)

\int \frac{3}{18 + 2 x ^{2}} d x

\frac{1}{2} arctan (\frac{1}{3} x) + C

Solution steps

\int \frac{3}{18 + 2 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{18 + 2 x ^{2}} d x

Apply Integral Substitution

= 3 \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

= 3 \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)

= 3 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{2}{3 2} x

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} arctan (\frac{1}{3} x) + C

\int_{- 2}^{1} 0 d x

\int_{0}^{4} (4 - x)^{2} d x

y^{^{''}} - (\frac{2}{x}) y^{^{'}} - (\frac{3}{x ^{2}}) y = 0

n = 0 \sum \infty cos (2 n)

r \to 1 lim (\frac{8 r + 1}{r + 3})

What is the integral of 3/(18+2x^2) ?
The integral of 3/(18+2x^2) is 1/2 arctan(1/3 x)+C