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Hint: Use the method of substitution and remember ${e^{\ln x}} = x$ and use that method to solve the question.

The given expression is$\int {{e^{2{x^2} + \ln x}}dx} = I$ . We are required to solve this integral.

We can observe that given integral is not in the standard form of integration so it cannot be solved directly. Hence we have to make changes in integral.

$I = \int {{e^{2{x^2} + \ln x}}dx} $

We can use the formula for splitting the expression using the formula of exponent i.e. ${a^{m + n}} = {a^m}{a^n}$ to change the integral into standard form to get an answer.

So, we can use it in the above integral and we will get

So we break the expression ${e^{2{x^2} + \ln x}}$ into ${e^{2{x^2}}}$ and ${e^{\ln x}}$ .

$I = \int {{e^{2{x^2}}}{e^{\ln x}}dx} $

We know that ${e^{\ln x}} = x$, so here we used it and we will get

$I = \int {{e^{2{x^2}}}xdx} $ …(1)

We need to convert the integral into standard form so, when integral contains a function and its derivative, then we can use the substitution method of integration to solve the integral.

Now we will use the method of substitution.

Then, let us consider ${e^{2{x^2}}} = t$

We have to differentiate with respect to $x$ on both sides.

We know that the formula of derivation.

That is $\dfrac{{d{e^x}}}{{dx}} = {e^x}$ and $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ .

And we will use the chain rule of derivation that $\dfrac{{d{e^{{x^n}}}}}{{dx}} = {e^{{x^n}}}\dfrac{{d{x^n}}}{{dx}}$ .

$\begin{gathered}

\dfrac{{d{e^{2{x^2}}}}}{{dx}} = \dfrac{{dt}}{{dx}} \\

{e^{2{x^2}}}\dfrac{{d2{x^2}}}{{dx}} = \dfrac{{dt}}{{dx}} \\

{e^{2{x^2}}}4x = \dfrac{{dt}}{{dx}} \\

\end{gathered} $

${e^{2{x^2}}}xdx = \dfrac{{dt}}{4}$ ..(2)

Using equation (2) and putting its value in equation (1) to convert the integral into standard form and we will get,

$I = \dfrac{1}{4}\int {dt} $

Now, we know that$\int {{x^n}dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}}$

Then we will get,

$I = \dfrac{t}{4}$

But we know that in the question, the variable used is $x$ so it is mandatory to get the final answer in terms of the given variable $x$.

Now we will replace the value of $t$ and represent the solution in terms of variable $x$ .

$I = \dfrac{{{e^{2{x^2}}}}}{4}$

Therefore, we will get our final answer that is

$\int {{e^{2{x^2} + \ln x}}dx} = \dfrac{{{e^{2{x^2}}}}}{4}$

Note: We can use an alternate way to solve the question using the method of integration by part in equation 1 but that method could get lengthy. Also whenever expressions consist of more than one kind of function and one of the functions is derivative of the other then we use a method of substitution. And whenever we use the substitution method of integration we must not forget to replace the substitute variable with the original variable $x$ used in the expression at the final answer.

The given expression is$\int {{e^{2{x^2} + \ln x}}dx} = I$ . We are required to solve this integral.

We can observe that given integral is not in the standard form of integration so it cannot be solved directly. Hence we have to make changes in integral.

$I = \int {{e^{2{x^2} + \ln x}}dx} $

We can use the formula for splitting the expression using the formula of exponent i.e. ${a^{m + n}} = {a^m}{a^n}$ to change the integral into standard form to get an answer.

So, we can use it in the above integral and we will get

So we break the expression ${e^{2{x^2} + \ln x}}$ into ${e^{2{x^2}}}$ and ${e^{\ln x}}$ .

$I = \int {{e^{2{x^2}}}{e^{\ln x}}dx} $

We know that ${e^{\ln x}} = x$, so here we used it and we will get

$I = \int {{e^{2{x^2}}}xdx} $ …(1)

We need to convert the integral into standard form so, when integral contains a function and its derivative, then we can use the substitution method of integration to solve the integral.

Now we will use the method of substitution.

Then, let us consider ${e^{2{x^2}}} = t$

We have to differentiate with respect to $x$ on both sides.

We know that the formula of derivation.

That is $\dfrac{{d{e^x}}}{{dx}} = {e^x}$ and $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ .

And we will use the chain rule of derivation that $\dfrac{{d{e^{{x^n}}}}}{{dx}} = {e^{{x^n}}}\dfrac{{d{x^n}}}{{dx}}$ .

$\begin{gathered}

\dfrac{{d{e^{2{x^2}}}}}{{dx}} = \dfrac{{dt}}{{dx}} \\

{e^{2{x^2}}}\dfrac{{d2{x^2}}}{{dx}} = \dfrac{{dt}}{{dx}} \\

{e^{2{x^2}}}4x = \dfrac{{dt}}{{dx}} \\

\end{gathered} $

${e^{2{x^2}}}xdx = \dfrac{{dt}}{4}$ ..(2)

Using equation (2) and putting its value in equation (1) to convert the integral into standard form and we will get,

$I = \dfrac{1}{4}\int {dt} $

Now, we know that$\int {{x^n}dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}}$

Then we will get,

$I = \dfrac{t}{4}$

But we know that in the question, the variable used is $x$ so it is mandatory to get the final answer in terms of the given variable $x$.

Now we will replace the value of $t$ and represent the solution in terms of variable $x$ .

$I = \dfrac{{{e^{2{x^2}}}}}{4}$

Therefore, we will get our final answer that is

$\int {{e^{2{x^2} + \ln x}}dx} = \dfrac{{{e^{2{x^2}}}}}{4}$

Note: We can use an alternate way to solve the question using the method of integration by part in equation 1 but that method could get lengthy. Also whenever expressions consist of more than one kind of function and one of the functions is derivative of the other then we use a method of substitution. And whenever we use the substitution method of integration we must not forget to replace the substitute variable with the original variable $x$ used in the expression at the final answer.

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