Courses
Courses for Kids
Free study material
Offline Centres
More

How do you write the value of $\int{x{{a}^{{{x}^{2}}+1}}dx}$?

seo-qna
Last updated date: 26th Feb 2024
Total views: 338.7k
Views today: 3.38k
Rank Predictor
Answer
VerifiedVerified
338.7k+ views
Hint: We first explain the terms $\dfrac{dy}{dx}$ where $y=f\left( x \right)$. We then need to integrate the equation once to find all the solutions of the integration. We take one arbitrary constant term for the integration. We use replacement of base value where $z={{x}^{2}}+1$ for $\int{x{{a}^{{{x}^{2}}+1}}dx}$. We use the formulas $\int{{{a}^{x}}dx}=\dfrac{{{a}^{x}}}{\log a}+c,\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.

Complete step-by-step solution:
We need to find the integral of $\int{x{{a}^{{{x}^{2}}+1}}dx}$.
We are going to change the base of the integral where we assume the new variable of $z={{x}^{2}}+1$.
We take the new base and differentiate the equation $z={{x}^{2}}+1$.
We know that the differentiated form of ${{x}^{2}}$ is $2x$ as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
Differentiating both sides with respect to $x$, we get
 \[\begin{align}
  & \dfrac{d}{dx}\left( z \right)=\dfrac{d}{dx}\left( {{x}^{2}}+1 \right) \\
 & \Rightarrow \dfrac{dz}{dx}=2x \\
\end{align}\]
Now we convert the differentiation into differential form where \[\dfrac{dz}{2}=xdx\].
Now we try to reform the main function of the integration where $x{{a}^{{{x}^{2}}+1}}dx={{a}^{{{x}^{2}}+1}}\left( xdx \right)$.
We now replace all those values with \[\dfrac{dz}{2}=xdx\] and $z={{x}^{2}}+1$ in the $\int{x{{a}^{{{x}^{2}}+1}}dx}$
$\begin{align}
  & \int{x{{a}^{{{x}^{2}}+1}}dx} \\
 & =\int{{{a}^{z}}\left( \dfrac{dz}{2} \right)} \\
 & =\dfrac{1}{2}\int{{{a}^{z}}dz} \\
\end{align}$
We simplify the integral equation by the formula $\int{{{a}^{x}}dx}=\dfrac{{{a}^{x}}}{\log a}+c$.
So, $\dfrac{1}{2}\int{{{a}^{z}}dz}=\dfrac{{{a}^{z}}}{2\log a}+c$.
We put the values where $z={{x}^{2}}+1$. We got $\dfrac{{{a}^{z}}}{2\log a}+c=\dfrac{{{a}^{{{x}^{2}}+1}}}{2\log a}+c$.
The integral of $\int{x{{a}^{{{x}^{2}}+1}}dx}$ is $\dfrac{{{a}^{{{x}^{2}}+1}}}{2\log a}+c$. Here $c$ is the integral constant.

Note: We can also solve those integration using the base change for ratio $z={{x}^{2}}$. In that case the sum gets complicated but the final solution would be the same. It is better to watch out for the odd power value in the ratios and take that as the change in the variable.