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Question: The value of \(\int {{e^{{{\tan }^{ - 1}}x}}} \dfrac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}d...

The value of ∫etan⁡−1x(1+x+x2)1+x2dx\int {{e^{{{\tan }^{ - 1}}x}}} \dfrac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}dx is
A. tan⁡−1x+C{\tan ^{ - 1}}x + C
B. etan⁡−1x+2x+C{e^{{{\tan }^{ - 1}}x}} + 2x + C
C. etan⁡−1x+C{e^{{{\tan }^{ - 1}}x}} + C
D. xetan⁡−1x+Cx{e^{{{\tan }^{ - 1}}x}} + C

Explanation

Solution

In the above question we have to integrate the given function. We can also see that in the expression we have a tangent which is a trigonometric ratio. So we will also apply the trigonometric formula to simplify this. And we should keep in mind that the final answer must be written in the original variable of integration. It should always have CC , known as the constant of integration or arbitrary constant. We should always add CC as the end of the solution.

Formula used:
1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta
Another formula that we will use here is
∫ex(f(x)+f′(x)dx=exf(x)+C\int {{e^x}(f(x) + f'(x)dx = {e^x}f(x) + C}

Complete answer:
Here we have ∫etan⁡−1x(1+x+x2)1+x2dx\int {{e^{{{\tan }^{ - 1}}x}}} \dfrac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}dx
Let us assume
tan⁡−1x=t{\tan ^{ - 1}}x = t
We will simplify this and it gives us
⇒1tan⁡x=t\Rightarrow \dfrac{1}{{\tan }}x = t
⇒x=tan⁡t\Rightarrow x = \tan t
Therefore let us write that
⇒dx11+x2=dt\Rightarrow dx\dfrac{1}{{1 + {x^2}}} = dt
We will now substitute these values in the expression and it can be written as ;
⇒I=∫et(1+tan⁡t+tan⁡2t)dt\Rightarrow I = \int {{e^t}(1 + \tan t + {{\tan }^2}t)dt}
Again here we can apply the formula that
1+tan⁡2t=sec⁡2t1 + {\tan ^2}t = {\sec ^2}t
By applying this formula we can write
=∫et(tan⁡t+sec⁡2t)dt= \int {{e^t}(\tan t + {{\sec }^2}t)dt}
Here we will apply the formula which states that
⇒∫ex(f(x)+f′(x)dx=exf(x)+C\Rightarrow \int {{e^x}(f(x) + f'(x)dx = {e^x}f(x) + C}
Therefore we can write that
=ettan⁡t+C= {e^t}\tan t + C
Now we know that we have assumed above
tan⁡−1x=t{\tan ^{ - 1}}x = t
So by substituting the value back in the equation, we have:
etan⁡−1xtan⁡(tan⁡−1x)+C{e^{{{\tan }^{ - 1}}x}}\tan ({\tan ^{ - 1}}x) + C
We know the basic trigonometric formula i.e
. ⇒tan⁡(tan⁡−1x)=x \Rightarrow \tan ({\tan ^{ - 1}}x) = x
By applying this we can write
⇒I=xetan⁡−1x+C\Rightarrow I = x{e^{{{\tan }^{ - 1}}x}} + C
Hence the correct option is (D) xetan⁡−1x+Cx{e^{{{\tan }^{ - 1}}x}} + C

Therefore, the correct option is D

Note: We should know that integration is the process of finding functions whose derivative is given and named anti-differentiation or integration. The function is called the anti-derivative or integral or primitive of the given function f(x)f(x) and CC is known as the constant of integration or the arbitrary constant. The function f(x)f(x) is called the integrand and dxdx is known as the element of integration.