What is the integral of 5/(x^2+6) ?

The integral of 5/(x^2+6) is 5/(sqrt(6))arctan(x/(sqrt(6)))+C

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integral of 5/(x^2+6)

\int \frac{5}{x ^{2} + 6} d x

\frac{5}{6} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{5}{x ^{2} + 6} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} + 6} d x

Apply Integral Substitution

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{x}{6}

= 5 \cdot \frac{1}{6} arctan (\frac{x}{6})

Simplify

5 \cdot \frac{1}{6} arctan (\frac{x}{6}) : \frac{5}{6} arctan (\frac{x}{6})

= \frac{5}{6} arctan (\frac{x}{6})

Add a constant to the solution

= \frac{5}{6} arctan (\frac{x}{6}) + C

cos^{2} (θ)

\int \frac{0 ^{1} ( x ^{2} - 2 x - 1 )}{x} d x

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

\int \frac{1}{3} x^{3} - 5 x d x

\int sec^{3} (x) cot (x) d x

What is the integral of 5/(x^2+6) ?
The integral of 5/(x^2+6) is 5/(sqrt(6))arctan(x/(sqrt(6)))+C