What is the integral of 6/(7+x^2) ?

The integral of 6/(7+x^2) is 6/(sqrt(7))arctan(x/(sqrt(7)))+C

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integral of 6/(7+x^2)

\int \frac{6}{7 + x ^{2}} d x

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

Solution steps

\int \frac{6}{7 + 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}{7 + x ^{2}} d x

Apply Integral Substitution

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

Substitute back

u = \frac{x}{7}

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

Simplify

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

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

Add a constant to the solution

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

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