Question

What is the integral of $\int{\arctan x}dx?$

Answer 1

Hint: We will use the integration by part to find the integral of the given function. Suppose that $f$ and $g$ are two functions of $x.$ Then the integral $\int{fg}dx$ can be found by using the integration by part. That is $\int{fg}dx=f\int{g}dx-\int{{f}'\int{g}dx}dx.$

Complete step by step solution:
We are asked to find the integral of $\int{\arctan x}dx.$
We know that the product of any number or function and $1$ is itself.
So, we will get $\arctan x\cdot 1=\arctan x.$
In this case, we can consider $f=\arctan x$ and $g=1.$
Now, by using this information, we can find the required integral with integration by parts.
Suppose that we have two functions of $x,$ namely $f$ and $g.$ Then the integral of their product can be found by $\int{fg}dx=f\int{gdx}-\int{{f}'\int{g}dx}dx.$ This rule of integration is called the integration by parts.
So, the given integral can be written as $\int{\arctan x}dx=\int{\arctan x\cdot 1}dx.$
Now, from this, we will get $\int{\arctan x}dx=\arctan x\int{1}dx-\int{\dfrac{d}{dx}\arctan x\int{1}dx}dx.$
We know that $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+{{x}^{2}}}$ and $\int{1}dx=x+c.$
So, we will get $\int{\arctan x}dx=\arctan x\cdot x-\int{\dfrac{1}{1+{{x}^{2}}}\cdot x}dx.$
That is, $\int{\arctan x}dx=x\arctan x-\int{\dfrac{x}{1+{{x}^{2}}}}dx.$
Let us put $1+{{x}^{2}}=u.$ Then $2xdx=du.$ This implies $xdx=\dfrac{du}{2}.$
So, we will get $\int{\dfrac{x}{1+{{x}^{2}}}dx=\int{\dfrac{1}{u}du.}}$
This will give us \[\int{\dfrac{x}{1+{{x}^{2}}}dx=\int{\dfrac{1}{u}du}=\ln \left| u \right|+c.}\]
And, we will get \[\int{\dfrac{x}{1+{{x}^{2}}}dx=\dfrac{1}{2}\int{\dfrac{1}{u}du}=\dfrac{1}{2}\ln \left| u \right|+c=\dfrac{1}{2}\ln \left| 1+{{x}^{2}} \right|+c.}\]
Therefore, we will get $\int{\arctan x}dx=x\arctan x-\dfrac{1}{2}\ln \left| 1+{{x}^{2}} \right|dx+C.$
Hence the integral is $\int{\arctan x}dx=x\arctan x-\dfrac{1}{2}\ln \left| 1+{{x}^{2}} \right|dx+C.$

Note: In indefinite integration, we will put a constant of integration $C.$ Also, we know the integral $\int{\dfrac{1}{x}}dx=\ln \left| x \right|+C.$ When we substitute variables for the function, the integration is called integration by substitution.