What is the integral of 6/(3+x^2) ?

The integral of 6/(3+x^2) is 2sqrt(3)arctan(x/(sqrt(3)))+C

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integral of 6/(3+x^2)

\int \frac{6}{3 + x ^{2}} d x

23 arctan (\frac{x}{3}) + C

Solution steps

\int \frac{6}{3 + x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{3 + x ^{2}} d x

Apply Integral Substitution

= 6 \cdot \int \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

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

Substitute back

u = \frac{x}{3}

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

Simplify

6 \cdot \frac{1}{3} arctan (\frac{x}{3}) : 23 arctan (\frac{x}{3})

= 23 arctan (\frac{x}{3})

Add a constant to the solution

= 23 arctan (\frac{x}{3}) + C

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