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How do you evaluate: $\int{\dfrac{1+x}{1+{{x}^{2}}}dx}$?

Last updated date: 09th Aug 2024
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Hint: We first break the functions in the numerator of $\int{\dfrac{1+x}{1+{{x}^{2}}}dx}$. We take the $\dfrac{dy}{dx}$ altogether. We integrate the functions separately. Then we take the addition to complete the problem. We also use the integral formula of $\int{\dfrac{dx}{1+{{x}^{2}}}}={{\tan }^{-1}}x+c,\int{\dfrac{dx}{x}}=\log \left| x \right|+c$.

Complete step-by-step solution:
We first break the integral function as $\int{\dfrac{1+x}{1+{{x}^{2}}}dx}=\int{\dfrac{1}{1+{{x}^{2}}}dx}+\int{\dfrac{x}{1+{{x}^{2}}}dx}$.
We integrate these two functions and take the addition to get the final solution.
We use the integral formula of $\int{\dfrac{dx}{1+{{x}^{2}}}}={{\tan }^{-1}}x$.
For the second part $\int{\dfrac{x}{1+{{x}^{2}}}dx}$, we are going to change the base of the integral where we assume the new variable of $z=1+{{x}^{2}}$.
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 $\int{\dfrac{x}{1+{{x}^{2}}}dx}$.
We now replace all those values with $\dfrac{dz}{2}=xdx$ and $z={{x}^{2}}+1$ in the $\int{\dfrac{x}{1+{{x}^{2}}}dx}$.
We simplify the integral equation by the formula $\int{\dfrac{dx}{x}}=\log \left| x \right|+c$.
$\int{\dfrac{x}{1+{{x}^{2}}}dx}=\int{\dfrac{dz}{2z}}=\dfrac{1}{2}\int{\dfrac{dz}{z}}=\dfrac{1}{2}\log \left| z \right|$
We put the values where $z={{x}^{2}}+1$. We got $\int{\dfrac{x}{1+{{x}^{2}}}dx}=\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|$.
We take the integral constant of $c$ as the final one. So, $\int{\dfrac{1+x}{1+{{x}^{2}}}dx}={{\tan }^{-1}}x+\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+c$
The final integral of $\int{\dfrac{1+x}{1+{{x}^{2}}}dx}$ is ${{\tan }^{-1}}x+\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+c$.

Note: We can also solve those integrations 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.