What is the integral of 1/(x^2+60) ?

The integral of 1/(x^2+60) is 1/(2sqrt(15))arctan(x/(2sqrt(15)))+C

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integral of 1/(x^2+60)

\int \frac{1}{x ^{2} + 60} d x

\frac{1}{2 15} arctan (\frac{x}{2 15}) + C

Solution steps

\int \frac{1}{x ^{2} + 60} d x

Apply Integral Substitution

= \int \frac{1}{2 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}{2 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}{2 15} arctan (u)

Substitute back

u = \frac{x}{2 15}

= \frac{1}{2 15} arctan (\frac{x}{2 15})

Add a constant to the solution

= \frac{1}{2 15} arctan (\frac{x}{2 15}) + C

x \to \infty lim (4 x e^{(\frac{1}{x})} - 4 x)

\int (2 y - \frac{1}{2 y}) d y

2 x^{2} - x^{4}

x \to - 3 lim (\frac{7}{( x + 3 ) ^{2}})

f (x) = \frac{3}{x}, at x = 2

What is the integral of 1/(x^2+60) ?
The integral of 1/(x^2+60) is 1/(2sqrt(15))arctan(x/(2sqrt(15)))+C