# Why :$\int\frac{dx}{a^2+x^2}= \frac 1a \tan^{-1}(\frac x a )$?

I have tried many times to get the way to proof the below identity using trigonometric substitution but i can't , Then Is there any simple way to know why $(1)$ is defined by this form ?

$$\int\frac{dx}{a^2+x^2}= \frac 1a \tan^{-1}(\frac x a )\tag{1}$$ ?

Note: For the specific question why exactly : $\int\frac{dx}{1+x^2}= \tan^{-1} x$ ?

• Substitute $x=au$, and use that $(\tan^{-1}(x))'=\frac{1}{1+x^2}$. – Botond May 21 '18 at 22:34
• $$\frac{dx}{a^2 + x^2} = \frac 1 {a^2} \frac{dx}{1 + (x/a)^2} = \frac 1 a \frac{d(x/a)}{1 + (x/a)^2}.$$ – user296602 May 21 '18 at 22:35
• I meant the proof for the standard integral : why for example : integral (1/x²+1)= arctan (x ) ? – zeraoulia rafik May 21 '18 at 22:35
• – Botond May 21 '18 at 22:36
• @zeraouliarafik That's a standard fact contained in most calculus textbooks, or in the Wikipedia article. – user296602 May 21 '18 at 22:36

I would substitute $x=a\tan \theta$ and use the identity $1+\tan^2\theta=\sec^2\theta$. Then

\begin{align} \int\frac{dx}{a^2+x^2}&=\int\frac{a\sec^2\theta}{a^2+a^2\tan^2\theta}\,d\theta \\ &=\int\frac{a\sec^2\theta}{a^2(1+\tan^2\theta)}\,d\theta \\ &=\frac{1}{a}\int\frac{\sec^2\theta}{\sec^2\theta}\,d\theta \\ &=\frac{1}{a}\int d\theta \\ &=\frac{1}{a}\theta+C \\ &=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C \end{align} Since, from our substitution, $$\theta=\arctan\left(\frac{x}{a}\right)$$

This integral $$\int\frac{dx}{a^2+x^2}= \frac 1a \tan^{-1}(\frac x a )\tag{1}$$

simply means that derivative of $$\frac 1a \tan^{-1}(\frac x a )\tag{1}$$ is $$\frac{1}{a^2+x^2}$$

You may differentiate to verify the result.

$$\int \frac{1}{a^2+x^2}dx$$

Apply Integral Substitution: $x=au$ $$=\int \frac{1}{a(u^2+1)}\,du$$ $$=\frac1a\int \frac{1}{u^2+1}\,du$$ Now use the formula:$\int \frac{1}{u^2+1}\,du=\arctan(u)$

$$=\frac1a \arctan(u)$$

Substitute back $u=\frac xa$

$$\frac1a \arctan(\frac xa)$$ $$\int \frac{1}{a^2+x^2}\,dx=\frac1a \arctan(\frac xa)+C$$