What is the integral of 1/(x^2+10) ?

The integral of 1/(x^2+10) is 1/(sqrt(10))arctan(x/(sqrt(10)))+C

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integral of 1/(x^2+10)

\int \frac{1}{x ^{2} + 10} d x

\frac{1}{10} arctan (\frac{x}{10}) + C

Solution steps

\int \frac{1}{x ^{2} + 10} d x

Apply Integral Substitution

= \int \frac{1}{10 ( 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}{10} \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}{10} arctan (u)

Substitute back

u = \frac{x}{10}

= \frac{1}{10} arctan (\frac{x}{10})

Add a constant to the solution

= \frac{1}{10} arctan (\frac{x}{10}) + C

x^{2} y^{^{''}} + x y^{^{'}} + 9 y = 0, y (1) = 0, y^{^{'}} (1) = 2.5

\int \frac{12 x}{e ^{2} - e ^{2 y^{2}}} d x

\int \frac{1}{( 4 y ) ^{2}} d y

f (x) = 6 sin (4 x)

(0, - 2), (3, 0)

What is the integral of 1/(x^2+10) ?
The integral of 1/(x^2+10) is 1/(sqrt(10))arctan(x/(sqrt(10)))+C