Question

How do you find the antiderivative of ${{\int{x\left( {{x}^{2}}+1 \right)}}^{100}}dx$ ?

Answer 1

We have been given a product of two polynomial functions whose antiderivative is to be found. From our prior knowledge of basic calculus, we know that the antiderivative is another name for the integral of the function. Thus, we shall find the integral of the given expression by applying simple substitution of the more complex term to simplify the integration process.

Complete step by step solution:
Given that ${{\int{x\left( {{x}^{2}}+1 \right)}}^{100}}dx$
In order to find the antiderivative of the function given and calculate its derivative, we shall substitute the denominator of the expression equal to some variable-t.
Let $t={{x}^{2}}+1$
Differentiating left hand side with respect to variable-t and right hand side with respect to variable-x, we get
$dt=\dfrac{d}{dx}\left( {{x}^{2}}+1 \right)$
We know by basic formulae of differentiation that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and differential of a constant term is zero.
$\Rightarrow dt=2xdx$
$\Rightarrow xdx=\dfrac{dt}{2}$
We shall now find the integral, $I$ of the given expression.
$\Rightarrow I={{\int{x\left( {{x}^{2}}+1 \right)}}^{100}}dx$
Now, substituting the values of variable-x and $xdx$ , we have
$\Rightarrow I={{\int{\left( {{x}^{2}}+1 \right)}}^{100}}xdx$
$\Rightarrow I={{\int{\left( t \right)}}^{100}}\dfrac{dt}{2}$
$\Rightarrow I=\dfrac{1}{2}{{\int{\left( t \right)}}^{100}}dt$
We know from the basic formulae of integration that $\int{{{u}^{n}}.du=\dfrac{{{u}^{n+1}}}{n+1}+C}$.
$\Rightarrow I=\dfrac{1}{2}\left( \dfrac{{{t}^{100+1}}}{100+1} \right)+C$
$\Rightarrow I=\dfrac{1}{2}\left( \dfrac{{{t}^{101}}}{101} \right)+C$
$\Rightarrow I=\dfrac{{{t}^{101}}}{202}+C$
Further, substituting the value of variable-t, we get
$\Rightarrow I=\dfrac{{{\left( {{x}^{2}}+1 \right)}^{101}}}{202}+C$
Therefore, the antiderivative of ${{\int{x\left( {{x}^{2}}+1 \right)}}^{100}}dx$ is $\dfrac{{{\left( {{x}^{2}}+1 \right)}^{101}}}{202}+C$.

Note: Since we do not have a direct formula of integration for the given polynomial which was raised to a power of 100, therefore, we integrated this function by applying simple substitution of $t={{x}^{2}}+1$. Here, one mistake which we could have made was substituting the value of $xdx$ equal to $dt$ instead of substituting the value of $xdx$ equal to $\dfrac{dt}{2}$. We must take care while making such substitutions as their ignorance could lead us to incorrect solutions.