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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{2 ( 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} \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} 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_{3 π}^{4 π} 2 sin (x) d x

\int \frac{1}{x 9 x ^{2} - 16} d x

\frac{\partial}{\partial y} (2 x^{2} + y^{2})

x^{2} \frac{d y}{d x} = x y + x^{2} e^{\frac{y}{x}}

y^{^{''}} + 6 y + 8 = 0, y (0) = 0, y^{^{'}} (0) = 12

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