Integral of $\frac{1}{1 + x^{2}}$

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

The integral of 11 + x^2 is ≤ft(x) + C.

The integral of $\frac{1}{1 + x^{2}}$ is $\arctan\left(x\right) + C$ .

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

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Graph of f(x) = 1 / (1 + x^2) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.78539816.

How to work it out

1
We want an antiderivative of f(x) = 1 / (1 + $x^{2}$ ).
$\int \frac{1}{1 + x^{2}}\,dx$
2
Method: standard integral: arctan.
3
That gives the antiderivative; add the constant C because any constant disappears when you differentiate:
$\int \frac{1}{1 + x^{2}}\,dx = \arctan\left(x\right) + C$
4
Check by differentiating the answer; you must get the original function back:
$\frac{d}{dx}\left[\arctan\left(x\right)\right] = \frac{1}{1 + x^{2}}$
5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:
$\int_{0}^{1} \frac{1}{1 + x^{2}}\,dx = \Big[\arctan\left(x\right)\Big]_{0}^{1} \approx 0.78539816$

LaTeX source of the formulas on this page

\[ \int \frac{1}{1 + x^{2}}\,dx = \arctan\left(x\right) + C \]
\[ \int \frac{1}{1 + x^{2}}\,dx \]
\[ \frac{d}{dx}\left[\arctan\left(x\right)\right] = \frac{1}{1 + x^{2}} \]
\[ \int_{0}^{1} \frac{1}{1 + x^{2}}\,dx = \Big[\arctan\left(x\right)\Big]_{0}^{1} \approx 0.78539816 \]

Definite integrals of the same function

Interval	Value
from 0 to 1	0.78539816
from 0 to 0.5	0.46364761

Questions

What is the integral of 1 / (1 +

x^{2}

The integral of $\frac{1}{1 + x^{2}}$ is $\arctan\left(x\right) + C$ .

Why do we add + C?

Every function F(x) + C has the same derivative, so the antiderivative is only defined up to a constant.

What is the integral of 1 / (1 +

x^{2}

) from 0 to 1?

About 0.78539816.

Integral of 11+x2\frac{1}{1 + x^{2}}1+x21​

How to work it out

Definite integrals of the same function

Questions

More integrals

Integral of $\frac{1}{1 + x^{2}}$