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Question: How do you write the value of \(\int{x{{a}^{{{x}^{2}}+1}}dx}\)?...

How do you write the value of ∫xax2+1dx\int{x{{a}^{{{x}^{2}}+1}}dx}?

Explanation

Solution

We first explain the terms dydx\dfrac{dy}{dx} where y=f(x)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=x2+1z={{x}^{2}}+1 for ∫xax2+1dx\int{x{{a}^{{{x}^{2}}+1}}dx}. We use the formulas ∫axdx=axlog⁡a+c,ddx(xn)=nxn−1\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 ∫xax2+1dx\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=x2+1z={{x}^{2}}+1.
We take the new base and differentiate the equation z=x2+1z={{x}^{2}}+1.
We know that the differentiated form of x2{{x}^{2}} is 2x2x as ddx(xn)=nxn−1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}.
Differentiating both sides with respect to xx, we get

& \dfrac{d}{dx}\left( z \right)=\dfrac{d}{dx}\left( {{x}^{2}}+1 \right) \\\ & \Rightarrow \dfrac{dz}{dx}=2x \\\ \end{aligned}$$ Now we convert the differentiation into differential form where $$\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 $\int{x{{a}^{{{x}^{2}}+1}}dx}$ $\begin{aligned} & \int{x{{a}^{{{x}^{2}}+1}}dx} \\\ & =\int{{{a}^{z}}\left( \dfrac{dz}{2} \right)} \\\ & =\dfrac{1}{2}\int{{{a}^{z}}dz} \\\ \end{aligned}$ 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.