What is the integral of 1/(3x^2+7) ?

The integral of 1/(3x^2+7) is 1/(sqrt(21))arctan(sqrt(3/7)x)+C

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integral of 1/(3x^2+7)

\int \frac{1}{3 x ^{2} + 7} d x

\frac{1}{21} arctan (\frac{3}{7} x) + C

Solution steps

\int \frac{1}{3 x ^{2} + 7} d x

Apply Integral Substitution

= \int \frac{1}{21 ( 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}{21} \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}{21} arctan (u)

Substitute back

u = \frac{3}{7} x

= \frac{1}{21} arctan (\frac{3}{7} x)

Combine same powers :

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

= \frac{1}{21} arctan (\frac{3}{7} x)

Add a constant to the solution

= \frac{1}{21} arctan (\frac{3}{7} x) + C

What is the integral of 1/(3x^2+7) ?
The integral of 1/(3x^2+7) is 1/(sqrt(21))arctan(sqrt(3/7)x)+C