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

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

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integral of 1/(u^2+1)

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

arctan (u) + C

Solution steps

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

Use the common integral:

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

= arctan (u)

Add a constant to the solution

= arctan (u) + C

\frac{d y}{d x} = \frac{1}{x}

\int_{0}^{1} (e^{x^{2}}) d x

\frac{x}{y} \frac{d y}{d x} + ln (y) = x + 1

\int \frac{3}{4} x^{4} d x

x \to 2 + lim (\frac{2 x + 1}{x - 2})