Solveeit Logo
Login

Question

Question: What is the integral of \(\int{\arctan x}dx?\)...

What is the integral of ∫arctan⁡xdx?\int{\arctan x}dx?

Explanation

Solution

We will use the integration by part to find the integral of the given function. Suppose that ff and gg are two functions of x.x. Then the integral ∫fgdx\int{fg}dx can be found by using the integration by part. That is ∫fgdx=f∫gdx−∫f′∫gdxdx.\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 ∫arctan⁡xdx.\int{\arctan x}dx.
We know that the product of any number or function and 11 is itself.
So, we will get arctan⁡x⋅1=arctan⁡x.\arctan x\cdot 1=\arctan x.
In this case, we can consider f=arctan⁡xf=\arctan x and g=1.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,x, namely ff and g.g. Then the integral of their product can be found by ∫fgdx=f∫gdx−∫f′∫gdxdx.\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 ∫arctan⁡xdx=∫arctan⁡x⋅1dx.\int{\arctan x}dx=\int{\arctan x\cdot 1}dx.
Now, from this, we will get ∫arctan⁡xdx=arctan⁡x∫1dx−∫ddxarctan⁡x∫1dxdx.\int{\arctan x}dx=\arctan x\int{1}dx-\int{\dfrac{d}{dx}\arctan x\int{1}dx}dx.
We know that ddxarctan⁡x=11+x2\dfrac{d}{dx}\arctan x=\dfrac{1}{1+{{x}^{2}}} and ∫1dx=x+c.\int{1}dx=x+c.
So, we will get ∫arctan⁡xdx=arctan⁡x⋅x−∫11+x2⋅xdx.\int{\arctan x}dx=\arctan x\cdot x-\int{\dfrac{1}{1+{{x}^{2}}}\cdot x}dx.
That is, ∫arctan⁡xdx=xarctan⁡x−∫x1+x2dx.\int{\arctan x}dx=x\arctan x-\int{\dfrac{x}{1+{{x}^{2}}}}dx.
Let us put 1+x2=u.1+{{x}^{2}}=u. Then 2xdx=du.2xdx=du. This implies xdx=du2.xdx=\dfrac{du}{2}.
So, we will get ∫x1+x2dx=∫1udu.\int{\dfrac{x}{1+{{x}^{2}}}dx=\int{\dfrac{1}{u}du.}}
This will give us ∫x1+x2dx=∫1udu=ln⁡∣u∣+c.\int{\dfrac{x}{1+{{x}^{2}}}dx=\int{\dfrac{1}{u}du}=\ln \left| u \right|+c.}
And, we will get ∫x1+x2dx=12∫1udu=12ln⁡∣u∣+c=12ln⁡∣1+x2∣+c.\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 ∫arctan⁡xdx=xarctan⁡x−12ln⁡∣1+x2∣dx+C.\int{\arctan x}dx=x\arctan x-\dfrac{1}{2}\ln \left| 1+{{x}^{2}} \right|dx+C.
Hence the integral is ∫arctan⁡xdx=xarctan⁡x−12ln⁡∣1+x2∣dx+C.\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.C. Also, we know the integral ∫1xdx=ln⁡∣x∣+C.\int{\dfrac{1}{x}}dx=\ln \left| x \right|+C. When we substitute variables for the function, the integration is called integration by substitution.