Integral of 1/(1 + x²)

Q: What is the integral of 1/(1 + x²)?

\int \frac{1}{x^{2} + 1}\,dx = \arctan\left(x\right) + C.

Q: What is the definite integral of 1/(1 + x²) from 0 to 1?

By the Fundamental Theorem of Calculus, \int_{0}^{1} \frac{1}{x^{2} + 1}\,dx = \Big[\arctan\left(x\right)\Big]_{0}^{1} = \frac{\pi}{4} \approx 0.785398.

Q: How can I check the answer?

Differentiate it: the derivative of \arctan\left(x\right) is \frac{1}{x^{2} + 1}. CalcViz checks every antiderivative this way before showing it.

\int \frac{1}{x^{2} + 1}\,dx = \arctan\left(x\right) + C

The derivative of arctan x, so the integral is arctan x + C. The total area under the whole curve is exactly π.

Step by step

Arctangent rule
∫1/(a² + u²) du = (1/a)·arctan(u/a).
$\int \frac{1}{x^{2} + 1}\,dx = \arctan\left(x\right)$

f(x) = 1/(1 + x²)
F(x) + C

Constant C = 0.0

Every choice of C gives a valid antiderivative, since shifting up or down doesn’t change the slope.

Open in the integral calculator

Questions students ask

What is the integral of 1/(1 + x²)?

$\int \frac{1}{x^{2} + 1}\,dx = \arctan\left(x\right) + C$ .

How do you integrate 1/(1 + x²)?

The derivative of arctan x, so the integral is arctan x + C. The total area under the whole curve is exactly π.

What is the definite integral of 1/(1 + x²) from 0 to 1?

By the Fundamental Theorem of Calculus, $\int_{0}^{1} \frac{1}{x^{2} + 1}\,dx = \Big[\arctan\left(x\right)\Big]_{0}^{1} = \frac{\pi}{4} \approx 0.785398$ .

How can I check the answer?

Differentiate it: the derivative of $\arctan\left(x\right)$ is $\frac{1}{x^{2} + 1}$ . CalcViz checks every antiderivative this way before showing it.