The integration of 1/(1+x^{2}) is equal to tan^{2} x. In this post, we will see how to integrate 1/(1+x^2).

## Integration of $\dfrac{1}{1+x^2}$

**Question:** What is the integration of $\dfrac{1}{1+x^2}$? That is, Find $\int \dfrac{1}{1+x^2} dx$

*Answer:* The integration of $\dfrac{1}{1+x^2}$ is $\tan^2 x$.

**Explanation:**

Let us substitute $x=\tan t$ $\cdots (\star)$

Differentiating with respect to $x$, we have

$1=\sec^2 t \dfrac{dt}{dx}$

$\Rightarrow dx=\sec^2 t dt$

Now, $\int \dfrac{1}{1+x^2} dx$ $=\int \dfrac{\sec^2 t}{1+\tan^2 t} dt$

$=\int \dfrac{\sec^2 t}{\sec^2 t} dt$ as we know that $\sec^2t=1+\tan^2 t$.

$=\int dt$ $=t+C$

$=\tan^{-1}x+C$ as we have from $(\star)$ that x=tan t

$\Rightarrow t=\tan^{-1} x$.

Thus, the integration of $\dfrac{1}{1+x^2}$ is $\tan^{-1}x+C$ where $C$ is an integration constant.

**Also Read:** **Integration of log(sinx) from 0 to pi/2**

## Definite integration of $\dfrac{1}{1+x^2}$

**Question:** Find the integral $\int_0^1 \dfrac{1}{1+x^2} dx$

*Answer:*

From the above, we know that $\int \dfrac{1}{1+x^2} dx=\tan^{-1}x$. Thus, we have

$\int_0^1 \dfrac{1}{1+x^2} dx$ $=[\tan^{-1} x]_0^1$

$=\tan^{-1}1-\tan^{-1}0$

$=\dfrac{\pi}{4}-0$ $=\dfrac{\pi}{4}$

**Also Read:**

Next, we find the following integral $\int \dfrac{\cos x}{1+\sin^2 x} dx$

*Solution:*

Put $\sin x =t$ $\therefore \cos x dx=dt$

Thus, $\int \dfrac{\cos x}{1+\sin^2 x} dx$ $=\int \dfrac{1}{1+t^2} dt$

$=\tan^{-1} t+C$ by the above formula.

$=\tan^{-1}(\sin x)+C$ where $C$ is an integral constant.

## FAQs

**Q1: What is the integration of 1/1+x^2?**

**Answer:** The integration of 1/1+x^2 is equal to tan^{-1}x+C.