
The value of $\int {{e^{{{\tan }^{ - 1}}x}}} \dfrac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}dx$ is
A. ${\tan ^{ - 1}}x + C$
B. ${e^{{{\tan }^{ - 1}}x}} + 2x + C$
C. ${e^{{{\tan }^{ - 1}}x}} + C$
D. $x{e^{{{\tan }^{ - 1}}x}} + C$
Answer
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Hint: 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 $C$ , known as the constant of integration or arbitrary constant. We should always add $C$ as the end of the solution.
Formula used:
$1 + {\tan ^2}\theta = {\sec ^2}\theta $
Another formula that we will use here is
$\int {{e^x}(f(x) + f'(x)dx = {e^x}f(x) + C} $
Complete answer:
Here we have $\int {{e^{{{\tan }^{ - 1}}x}}} \dfrac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}dx$
Let us assume
${\tan ^{ - 1}}x = t$
We will simplify this and it gives us
$ \Rightarrow \dfrac{1}{{\tan }}x = t$
$ \Rightarrow x = \tan t$
Therefore let us write that
$ \Rightarrow dx\dfrac{1}{{1 + {x^2}}} = dt$
We will now substitute these values in the expression and it can be written as ;
$ \Rightarrow I = \int {{e^t}(1 + \tan t + {{\tan }^2}t)dt} $
Again here we can apply the formula that
$1 + {\tan ^2}t = {\sec ^2}t$
By applying this formula we can write
$ = \int {{e^t}(\tan t + {{\sec }^2}t)dt} $
Here we will apply the formula which states that
$ \Rightarrow \int {{e^x}(f(x) + f'(x)dx = {e^x}f(x) + C} $
Therefore we can write that
$ = {e^t}\tan t + C$
Now we know that we have assumed above
${\tan ^{ - 1}}x = t$
So by substituting the value back in the equation, we have:
${e^{{{\tan }^{ - 1}}x}}\tan ({\tan ^{ - 1}}x) + C$
We know the basic trigonometric formula i.e
. $ \Rightarrow \tan ({\tan ^{ - 1}}x) = x$
By applying this we can write
$ \Rightarrow I = x{e^{{{\tan }^{ - 1}}x}} + C$
Hence the correct option is (D) $x{e^{{{\tan }^{ - 1}}x}} + C$
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 $f(x)$ and $C$ is known as the constant of integration or the arbitrary constant. The function $f(x)$ is called the integrand and $dx$ is known as the element of integration.
Formula used:
$1 + {\tan ^2}\theta = {\sec ^2}\theta $
Another formula that we will use here is
$\int {{e^x}(f(x) + f'(x)dx = {e^x}f(x) + C} $
Complete answer:
Here we have $\int {{e^{{{\tan }^{ - 1}}x}}} \dfrac{{\left( {1 + x + {x^2}} \right)}}{{1 + {x^2}}}dx$
Let us assume
${\tan ^{ - 1}}x = t$
We will simplify this and it gives us
$ \Rightarrow \dfrac{1}{{\tan }}x = t$
$ \Rightarrow x = \tan t$
Therefore let us write that
$ \Rightarrow dx\dfrac{1}{{1 + {x^2}}} = dt$
We will now substitute these values in the expression and it can be written as ;
$ \Rightarrow I = \int {{e^t}(1 + \tan t + {{\tan }^2}t)dt} $
Again here we can apply the formula that
$1 + {\tan ^2}t = {\sec ^2}t$
By applying this formula we can write
$ = \int {{e^t}(\tan t + {{\sec }^2}t)dt} $
Here we will apply the formula which states that
$ \Rightarrow \int {{e^x}(f(x) + f'(x)dx = {e^x}f(x) + C} $
Therefore we can write that
$ = {e^t}\tan t + C$
Now we know that we have assumed above
${\tan ^{ - 1}}x = t$
So by substituting the value back in the equation, we have:
${e^{{{\tan }^{ - 1}}x}}\tan ({\tan ^{ - 1}}x) + C$
We know the basic trigonometric formula i.e
. $ \Rightarrow \tan ({\tan ^{ - 1}}x) = x$
By applying this we can write
$ \Rightarrow I = x{e^{{{\tan }^{ - 1}}x}} + C$
Hence the correct option is (D) $x{e^{{{\tan }^{ - 1}}x}} + C$
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 $f(x)$ and $C$ is known as the constant of integration or the arbitrary constant. The function $f(x)$ is called the integrand and $dx$ is known as the element of integration.
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