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Integrate $\int {{x^{{x^2} + 1}}\left( {2\ln x + 1} \right)} dx$.

Last updated date: 17th Sep 2024
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Hint: These type of question can be solved using the derivation, and we can use derivation of multiplication of two variables i.e., $\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$and derivative of log functions i.e., $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$., and also integrating formula i.e., $\int {dx} = x + C$.

Given question is $\int {{x^{{x^2} + 1}}\left( {2\ln x + 1} \right)} dx$.
We have to find the integration of the given question and this can be solved by using derivation, and the integration formulas.
Let${x^{{x^2}}} = t$,
Now applying log on both sides we get,
$\Rightarrow$$\ln {x^{{x^2}}} = \ln t$
As using logarithms identity i.e.,$\ln {x^2} = 2\ln x$,
$\Rightarrow$${x^2}\ln x = \ln t$,
Now differentiating on both sides we get,
$\Rightarrow$$\dfrac{d}{{dx}}{x^2}\ln x = \dfrac{d}{{dx}}\ln t$,
Now using derivation of multiplication of two variables i.e., $\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$, we get,
$\Rightarrow$$\ln x\dfrac{d}{{dx}}{x^2} + {x^2}\dfrac{d}{{dx}}\ln x = \dfrac{1}{t}dt$,
Now simplifying left hand side we get,
$\Rightarrow \ln x\left( {2x} \right) + {x^2}\left( {\dfrac{1}{x}} \right) = \dfrac{1}{t}dt$ ,
Again simplifying on left side we get,
$\Rightarrow 2x\ln x + x = \dfrac{1}{t}dt$,
Now taking t to L.H.S we get,
$\Rightarrow tx\left( {2\ln x + 1} \right) = dt$,
Now substituting t value in the above equation we get,
$\Rightarrow {x^{{x^2}}}x\left( {2\ln x + 1} \right) = dt$,
Now using exponential formulas i.e., ${x^a} \cdot {x^b} = {x^{a + b}}$we get,
$\Rightarrow {x^{{x^2} + 1}}\left( {2\ln x + 1} \right) = dt$,
Now integrating on both sides we get,
$\Rightarrow \int {{x^{{x^2} + 1}}\left( {2\ln x + 1} \right) = \int {dt} }$,
Now using integrating formulas we get,
$\Rightarrow {x^{{x^2} + 1}}\left( {2\ln x + 1} \right) = t + C$
Now substituting the value of t=${x^{{x^2}}}$ in the above equation we get,
$\Rightarrow {x^{{x^2} + 1}}\left( {2\ln x + 1} \right) = {x^{{x^2}}} + C$
Thus the integrated value of given equation will be ${x^{{x^2}}} + C$i.e.,
$\therefore {x^{{x^2} + 1}}\left( {2\ln x + 1} \right) = {x^{{x^2}}} + C$.

The integrated value of given equation is ${x^{{x^2}}} + C$

So, $\therefore {x^{{x^2} + 1}}\left( {2\ln x + 1} \right) = {x^{{x^2}}} + C$.

Note:
In these types of questions as we are using both derivation and integration formulas, students should not get confused as where to use the formulas and which one to use as there are many formulas in both derivations and integrations. Some of the important formulas that can be used are given below:
$\Rightarrow$$\dfrac{d}{{dx}}x = 1$
$\Rightarrow$$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
$\Rightarrow$$\dfrac{d}{{dx}}cx = c$
$\Rightarrow$$\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
$\Rightarrow$$\dfrac{d}{{dx}}\dfrac{u}{v} = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}$
$\Rightarrow$$\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$
$\Rightarrow$$\int {dx} = x + C$
$\Rightarrow$$\int {\dfrac{1}{x}dx = \ln x} + C$
$\Rightarrow$$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$