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I've been trying to integrate that function, but it looks like I'm missing something, so can any one please show me how to integrate $e^\sqrt{x}$, in order to correct my procedure. Thanks

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If $x=u^2$, then $\mathrm dx=\dots$ – J. M. Sep 26 '11 at 21:53
Hint: substitute $u = \sqrt{x}$. – Robert Israel Sep 26 '11 at 21:54
No is not a homework I'd tried to do the same substitution that is shown down in the first answer, but then I don't know what to do then, since there is a product in the integrand. – Davynch0 Sep 26 '11 at 22:13
up vote 10 down vote accepted

You're going to want to start with a $u$-substitution $u=\sqrt{x}$. Notice then that $u^2=x$, so $2udu=dx$. Then $$ \int e^\sqrt{x}dx=2\int e^u u\ du. $$

You can then proceed by integration by parts.

To get started, you can set $w=u$ and $dv=e^u\ du$, and use the fact that $\int w\ dv= wv-\int v\ dw$. Lastly, don't forget to substitute back to get the integral in terms of $x$.

Continuing, $dw=du$ and $v=e^u$. Then $$ \int w\ dv= wv-\int v\ dw = e^u u-\int e^u\ du=e^u u-e^u. $$ So $$ \int e^\sqrt{x}dx=2\int e^u u\ du=2e^uu-2e^u+C=2e^\sqrt{x}(\sqrt{x}-1)+C $$ after back substituting $u=\sqrt{x}$ and adding a possible constant term.

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I don't know how to integrate by parts, can't it be solved by the substitution rule at all? – Davynch0 Sep 26 '11 at 22:07
@Dav Since you say it is not homework, I've tried to add a more complete solution. Hopefully it's clearer. – yunone Sep 26 '11 at 22:20
thanks a lot @yunone – Davynch0 Sep 26 '11 at 22:21
Not homework, and you don't know integration by parts? Maybe that is why it is not (yet) homework. Maybe it WILL be homework later, after you do learn integration by parts! – GEdgar Sep 27 '11 at 0:12
@Davyncho: It really is (after the natural substitution) a standard integration by parts. But if we insist on substitution alone, we can (I am joking, but it is true) use $u=(\sqrt{x}-1)e^\sqrt{x}$. – André Nicolas Sep 27 '11 at 1:37

You can integrate $e^\sqrt{x}$ by substituting any single variable in place of $\sqrt{x}$.

Since, $e^\sqrt{x}$ does not look like $e^{x}$ due to the presence of square root, therefore it can not be integrated by normal methods.

$\int e^\sqrt{x} \,dx\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(1)$

To integrate $e^\sqrt{x}$, you should substitute any variable, say $t$ in place of $\sqrt{x}.$

Now, $e^\sqrt{x}$ becomes $e^t$ which is in the form of $e^x$, so it can be integrated easily.

But, when you substitute $t=$$\sqrt{x}$, then equation $(1)$ becomes :

$\int e^t \,dx$

and $t$ can not be integrated with respect to $x.$

So, replace $dx$ by evaluating its value as –



On differentiation


Thus, $\int e^\sqrt{x}\,dx=\int e^t \cdot 2t\,dt$

$=2\int e^t \cdot t\,dt \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ...(2)$

Now, you have to integrate a product of two functions, so you need the formula for integration by parts.

$\int u\,dv=uv - \int v\,du$

Here, $u$ and $v$ should be assumed by following the order of ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric and Exponential) from left to right.

For example : If you have a product of two terms, say $x \cdot sec^2 x \,dx,$ then according to the order of ILATE, algebraic term $(x)$ comes before the trigonometric term $(sec^2 x).$

Therefore, you should consider the algebraic term $(x)$ as $u$ and the trigonometric term $(sec^2 x\,dx)$ as $dv$.

Now, going back to the original product i.e., $e^t \cdot t\,dt$, we find that $t$ is an algebraic term which comes before the exponential term, $e^t.$

So, let $u=t$ and $dv= e^t\,dt$

Now, you need the values of $v$ and $du$ for the formula.

$\int u\,dv=uv–\int v \,du$

Therefore, differentiate $u=t$ and integrate $dv=e^t\,dt$

$du=dt$ and $v=e^t$

Put all these values in the formula –

$\int e^t \cdot t \,dt=t \cdot e^t – \int e^t\,dt$

$=t \cdot e^t – e^t$

Now, from equation (2)

$\int e^\sqrt{x}\,dx = 2 \int e^t \cdot t \,dt$

$=2 (t \cdot e^t – e^t )+c$

Take $e^t$ out of the parenthesis

$=2 e^t (t–1)+c$

Put the value of $t$ to get the answer in terms of $x.$

So, $\int e^\sqrt{x}\,dx = 2 e^\sqrt{x} (\sqrt{x}–1)+c$

Here, $c$ is the constant.

This, is the integral of $\int e^\sqrt{x} \,dx.$ enter image description here

You can also integrate different types of functions using substitution method, which otherwise can’t be integrated by normal methods.

Let’s integrate $cos^3 x$ to understand the substitution rule better.

$\int cos^3 x\,dx$

$cos^3 x$ can not be integrated directly. So, we need to factorize it as :

$\int cos^2 x \cdot cos x\,dx$

Now, if we substitute $u=cos x$

then on differentiation, we get :

$du=–sin x\,dx$

But, we don’t have $–sin x$ in both of the factors .

i.e., $cos^2 x, cos x$

So, we can’t substitute $u=cos x$

For substitution, we should have both the functions, $sin x$ and $cos x$

$\because\, sin^2 x + cos^2 x = 1$

Subtracting $sin^2 x$ from both the sides

$\Rightarrow sin^2 x + cos^2 x – sin^2 x = 1 – sin^2 x$

$\Rightarrow sin^2 x – sin^2 x + cos^2 x = 1 – sin^2 x$

$cos^2x = 1 – sin^2 x$

So, we can replace $cos^2 x$ by $1 – sin^2 x$

Then $\int cos^2 x \cdot cos x\,dx$ becomes, $\int (1–sin^2 x) \cdot cos x\, dx$

Now, we can substitute $u=sin x$

Thus, $\int (1–sin^2 x)\cdot cos x \,dx$ becomes,

$\int (1–u^2) cos x \,dx$

Differentiate $u=sin x$

$du =cos x \,dx$

So, $\int (1–u^2) cos x \,dx$

$= \int (1–u^2) du$

Now, it is easy to integrate

$=\int 1 \cdot du – \int u^2 \,du$

$=u –\frac{u^{2+1}}{2+1} +c$

$=u –\frac{u^3}{3} +c$

put the value of $u $

Then, $\int cos^3 x \,dx = sin x – \frac{sin^3 x}{3} +c$


$\int cos^3 x \,dx = sin x(1 – \frac{sin^2 x}{3}) +c $

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