# MX2 Integration Marathon 2021 (1 Viewer)

#### CM_Tutor

##### Moderator
Moderator
more integrals

$\bg_white \int \frac{\sin{\sqrt{x}+a}(e^{\sqrt{x}})}{\sqrt{x}}$
Do you mean

\bg_white \begin{align*} \int \frac{e^{\sqrt{x}}\sin{(\sqrt{x}+a)}}{\sqrt{x}}\ dx &= 2\int{e^u\sin(u+a)\ du} \qquad \text{where u=\sqrt{x}} \\ &= e^{\sqrt{x}}\left[\sin(\sqrt{x}+a)-\cos(\sqrt{x}+a)\right] + C \end{align*}

#### idkkdi

##### Well-Known Member
Do you mean

\bg_white \begin{align*} \int \frac{e^{\sqrt{x}}\sin{(\sqrt{x}+a)}}{\sqrt{x}}\ dx &= 2\int{e^u\sin(u+a)\ du} \qquad \text{where u=\sqrt{x}} \\ &= e^{\sqrt{x}}\left[\sin(\sqrt{x}+a)-\cos(\sqrt{x}+a)\right] + C \end{align*}
yes

#### CM_Tutor

##### Moderator
Moderator
At first, I thought you meant a much more difficult problem:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx$

#### idkkdi

##### Well-Known Member
At first, I thought you meant a much more difficult problem:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx$
@Qeru here ya go m8, im out

#### Qeru

##### Well-Known Member
At first, I thought you meant a much more difficult problem:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx$
Surely theres no elementary anti-derivative to this

#### CM_Tutor

##### Moderator
Moderator
Surely theres no elementary anti-derivative to this
None that I am aware of.

#### CM_Tutor

##### Moderator
Moderator
Though I can give a 1 line answer:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx = F(x) \quad \text{where} \quad \frac{dF}{dx} = \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}$

#### idkkdi

##### Well-Known Member
Though I can give a 1 line answer:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx = F(x) \quad \text{where} \quad \frac{dF}{dx} = \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}$
... that isnt the integral.

am i missing something? this doesn't look like a one line answer.

#### Qeru

##### Well-Known Member
... that isnt the integral.

am i missing something? this doesn't look like a one line answer.
He basically defined that integral as a function similar to the error function or Si(x) etc.

#### idkkdi

##### Well-Known Member
He basically defined that integral as a function similar to the error function or Si(x) etc.
what. it looks like he diffed the integral to get what's the integrand to me.

#### Qeru

##### Well-Known Member
Ok anyone want to try this (stumped me lol):

$\bg_white \int \frac{\sqrt{\sin{\sqrt{x}}}\cos{\sqrt{x}}}{1+x^2} dx$
Have you tried this @CM_Tutor I'm pretty sure theres an anti-derivative it's just really tricky.

#### CM_Tutor

##### Moderator
Moderator
Ok anyone want to try this (stumped me lol):

$\bg_white \int \frac{\sqrt{\sin{\sqrt{x}}}\cos{\sqrt{x}}}{1+x^2} dx$
I don't see an obvious way to approach this. Integral calculator can't find an answer in elementary functions, which suggests that none of the common approaches will work. What makes you confident that there is an answer with the realm of MX2 possibilities?

#### CM_Tutor

##### Moderator
Moderator
This is much tougher than I first thought as you keep running into the exponential integral, $\bg_white \text{Ei}(x)$ (or a transform of it), being integrals such as

$\bg_white \int{\frac{e^x}{x}}\ dx$

Transforms include the denominator being a linear function of $\bg_white x$ or the numerator as $\bg_white e^{-x}$. Properly, the exponential integral is defined (over the complex plane) as

$\bg_white \text{Ei}(x)=\int_{-x}^{\infty}{\frac{e^{-t}}{t}}\ dt=\int_{-\infty}^{x}{\frac{e^t}{t}}\ dt$

and it cannot be expressed in elementary functions.

However, there is a closed form for this integral:

$\bg_white \int{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{xe^x}{x+2} +C \text{, for some constant C}$

$\bg_white \int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{e}{3}$

For those who want to try to solve the problem yourselves, you need to rewrite the integrand without creating terms of the form

$\bg_white \frac{Ae^x}{(mx+b)^n$

where $\bg_white A$, $\bg_white m$, and $\bg_white b$, are constants and, in this case, $\bg_white n=1$ or $\bg_white n=2$.

You will need something of the form:

$\bg_white \left(\frac{f(x)}{x+2}+\frac{g(x)}{(x+2)^2}\right)e^x$

where $\bg_white f(x)$ and $\bg_white g(x)$ are both of degree 1.

\bg_white \begin{align*} I&=\int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx \\ &=\int_0^1{\frac{x^2+3x + 2 - x}{(x+2)^2}e^x}\ dx \\ &=\int_0^1{\left(\frac{(x+1)(x+2) - x}{(x+2)^2}\right)e^x}\ dx \\ &=\int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \\ &=J - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \qquad \text{where } J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx \end{align*}

We know to leave $\bg_white J$ alone as it is clearly a problem:

$\bg_white J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx = \int_0^1{\frac{(x + 2 - 1)e^x}{x+2}}\ dx = \int_0^1{e^x - \frac{e^x}{x+2}}\ dx$

i.e. $\bg_white J$ is a constant plus (or minus) the exponential integral.

\bg_white \begin{align*} I&=J - \int_0^1{xe^x}\ d\left(\frac{-1}{x+2}\right) \\ &=J - \left(\left[\frac{-xe^x}{x+2}\right]_0^1 + \int_0^1{\frac{-1}{x+2}}\ d(xe^x)\right) \\ &=J - \left[\frac{-1 \times e^1}{1+2} - \frac{-0 \times e^0}{0+2}\right] - \int_0^1{\frac{e^x + xe^x}{x+2}}\ dx \\ &=J + \frac{e}{3} - 0 - \int_0^1{\frac{e^x(x+1)}{x+2}}\ dx \\ &=J + \frac{e}{3} - J \\ &= \frac{e}{3} \end{align*}

#### s97127

##### Member
This is much tougher than I first thought as you keep running into the exponential integral, $\bg_white \text{Ei}(x)$ (or a transform of it), being integrals such as

$\bg_white \int{\frac{e^x}{x}}\ dx$

Transforms include the denominator being a linear function of $\bg_white x$ or the numerator as $\bg_white e^{-x}$. Properly, the exponential integral is defined (over the complex plane) as

$\bg_white \text{Ei}(x)=\int_{-x}^{\infty}{\frac{e^{-t}}{t}}\ dt=\int_{-\infty}^{x}{\frac{e^t}{t}}\ dt$

and it cannot be expressed in elementary functions.

However, there is a closed form for this integral:

$\bg_white \int{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{xe^x}{x+2} +C \text{, for some constant C}$

$\bg_white \int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{e}{3}$

For those who want to try to solve the problem yourselves, you need to rewrite the integrand without creating terms of the form

$\bg_white \frac{Ae^x}{(mx+b)^n$

where $\bg_white A$, $\bg_white m$, and $\bg_white b$, are constants and, in this case, $\bg_white n=1$ or $\bg_white n=2$.

You will need something of the form:

$\bg_white \left(\frac{f(x)}{x+2}+\frac{g(x)}{(x+2)^2}\right)e^x$

where $\bg_white f(x)$ and $\bg_white g(x)$ are both of degree 1.

\bg_white \begin{align*} I&=\int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx \\ &=\int_0^1{\frac{x^2+3x + 2 - x}{(x+2)^2}e^x}\ dx \\ &=\int_0^1{\left(\frac{(x+1)(x+2) - x}{(x+2)^2}\right)e^x}\ dx \\ &=\int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \\ &=J - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \qquad \text{where } J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx \end{align*}

We know to leave $\bg_white J$ alone as it is clearly a problem:

$\bg_white J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx = \int_0^1{\frac{(x + 2 - 1)e^x}{x+2}}\ dx = \int_0^1{e^x - \frac{e^x}{x+2}}\ dx$

i.e. $\bg_white J$ is a constant plus (or minus) the exponential integral.

\bg_white \begin{align*} I&=J - \int_0^1{xe^x}\ d\left(\frac{-1}{x+2}\right) \\ &=J - \left(\left[\frac{-xe^x}{x+2}\right]_0^1 + \int_0^1{\frac{-1}{x+2}}\ d(xe^x)\right) \\ &=J - \left[\frac{-1 \times e^1}{1+2} - \frac{-0 \times e^0}{0+2}\right] - \int_0^1{\frac{e^x + xe^x}{x+2}}\ dx \\ &=J + \frac{e}{3} - 0 - \int_0^1{\frac{e^x(x+1)}{x+2}}\ dx \\ &=J + \frac{e}{3} - J \\ &= \frac{e}{3} \end{align*}
That's correct. what's the hardest integration problem that you have solved?

#### stupid_girl

##### Active Member
I know @vernburn has correctly split the interval. However, periodicity actually made the calculation easier rather than causing trouble.

Depending on how you express the antiderivative, the constant of integration may be different.