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

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

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integral of 1/(1^2+2x^2)

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

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

Solution steps

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

Simplify

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Add a constant to the solution

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

\int \frac{x ^{3} - 5 x ^{2} + 3 x + 14}{x - 3} d x

f (x) = e^{- 3 x}, (0, 1)

\frac{d x}{d t} = 5 x (6 - x)

\frac{d y}{d x} + 3 y = 50

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{y ^{5}}

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