What is the integral of 1/(3x^2+5) ?

The integral of 1/(3x^2+5) is 1/(sqrt(15))arctan(sqrt(3/5)x)+C

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integral of 1/(3x^2+5)

\int \frac{1}{3 x ^{2} + 5} d x

\frac{1}{15} arctan (\frac{3}{5} x) + C

Solution steps

\int \frac{1}{3 x ^{2} + 5} d x

Apply Integral Substitution

= \int \frac{1}{15 ( 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}{15} \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}{15} arctan (u)

Substitute back

u = \frac{3}{5} x

= \frac{1}{15} arctan (\frac{3}{5} x)

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{15} arctan (\frac{3}{5} x)

Add a constant to the solution

= \frac{1}{15} arctan (\frac{3}{5} x) + C

x^{2} y^{^{''}} - 7 x y^{^{'}} + 16 y = 0

y = - 3 x^{3} + x^{2} + 10 x, x \geq 0

y^{^{''}} - 2 y^{^{'}} - 3 y = x e^{x}

e^{\frac{4}{5} x}

\frac{d}{d x} (x + 1 - 2)

What is the integral of 1/(3x^2+5) ?
The integral of 1/(3x^2+5) is 1/(sqrt(15))arctan(sqrt(3/5)x)+C