MHB Integration by Parts: Solve $$\frac{xe^{2x}}{(1+2x)^2}$$

[ineedhelpnow]

Im supposed to use integration by parts for this problem but i understand how to.

$$\int \ \frac{xe^{2x}}{(1+2x)^2},dx$$

 

[evinda]

Im supposed to use integration by parts for this problem but i understand how to.

$$\int \ \frac{xe^{2x}}{(1+2x)^2},dx$$

You can do it like that:
$$ \int \frac{xe^{2x}}{(1+2x)^2}=\int \frac{1}{2} x e^{2x} (\frac{-1}{1+2x})'dx=\frac{-xe^{2x}}{2(1+2x)}+\frac{1}{2} \int \frac{(e^{2x}+2xe^{2x})}{1+2x}dx \\ =\frac{-xe^{2x}}{2(1+2x)}+\frac{1}{2} \int \frac{e^{2x}(1+2x)}{1+2x}dx=\frac{-xe^{2x}}{2(1+2x)}+\frac{1}{2} \int e^{2x}dx=\frac{-xe^{2x}}{2(1+2x)}+\frac{1}{2} \frac{e^{2x}}{2}+c=\frac{-2xe^{2x}+(1+2x)e^{2x}}{4(1+2x)}+c=\frac{e^{2x}}{4+8x}+c$$

 

[ineedhelpnow]

are these the right methods for these problems?

$\int_{0}^{1/8} \ ytan(8y),dy$ Integration By Parts

$\int \ 6(sec(x))^4,dx$ Reduction

$\int \ \frac{7x^2+x+24}{x^3+4x},dx$ Partial Fractions

$\int \ \frac{e^{2x}}{6e^x+4},dx$ no idea

 

[evinda]

are these the right methods for these problems?$\int \ \frac{7x^2+x+24}{x^3+4x},dx$ Partial Fractions

$\int \ \frac{e^{2x}}{6e^x+4}dx$ no idea

For the integral $\int \ \frac{7x^2+x+24}{x^3+4x},dx$ ,you are right that you have to use the method Partial Fractions,that you have mentioned.

For the integral $\int \ \frac{e^{2x}}{6e^x+4}dx$ ,set $e^x=u$.
Then $e^x dx=du$

Therefore:

$$\int \frac{e^{2x}}{8e^x+4}dx=\int \frac{u}{8u+4}du=\int \frac{8u}{8(8u+4)}du=\frac{1}{8} \int \frac{8u+4-4}{8u+4}du=\frac{1}{8} \int (1-\frac{4}{8u+4})du=\frac{1}{8}[u- \frac{1}{2} \ln{|2u+1|}]+c=\frac{1}{8}(e^x-\frac{1}{2} \ln{(2e^x+1)})+c=\frac{e^x}{8}-\frac{1}{16} \ln{(2e^x+1)}+c$$

 

[ineedhelpnow]

thank you so much! your explanation made it totally clear.for the 2nd one i listed i used reduction and it worked but how do i go about the first one?

 

[ineedhelpnow]

For the integral $\int \ \frac{e^{2x}}{6e^x+4}dx$ ,set $e^x=u$.
Then $e^x dx=du$

Therefore:

$$\int \frac{e^{2x}}{8e^x+4}dx=\int \frac{u}{8u+4}du=\int \frac{8u}{8(8u+4)}du=\frac{1}{8} \int \frac{8u+4-4}{8u+4}du=\frac{1}{8} \int (1-\frac{4}{8u+4})du=\frac{1}{8}[u- \frac{1}{2} \ln{|2u+1|}]+c=\frac{1}{8}(e^x-\frac{1}{2} \ln{(2e^x+1)})+c=\frac{e^x}{8}-\frac{1}{16} \ln{(2e^x+1)}+c$$

you plugged in u for the numerator on the first step but originally it was $e^{2x}$ not $e^{x}$ and did you mean 6 (instead of 8)?

 

[evinda]

you plugged in u for the numerator on the first step but originally it was $e^{2x}$ not $e^{x}$ and did you mean 6 (instead of 8)?

Yes,in the first step it is actually $\int \frac{u}{6u+4} du$,because:

$$e^x=u \Rightarrow e^xdx=du$$

Therefore:

$$\int \frac{e^{2x}}{6e^x+4}dx= \int \frac{e^x \cdot e^x dx}{6e^x+4}dx= \int \frac{u du}{6u+4}$$

Oh,and sorry for calculating $\frac{e^{2x}}{8e^x+4}dx$,instead of $\frac{e^{2x}}{6e^x+4}dx$ ! I accidentally wrote $8$ instead of $6$.. (Blush)

 

[ineedhelpnow]

ooohhh i get what you did. makes sense. is there any other way to approach this problem or is this the only way? (i have a test the day after tomorrow so i need to REALLY understand this stuff)

 

[evinda]

thank you so much! your explanation made it totally clear.for the 2nd one i listed i used reduction and it worked but how do i go about the first one?

For the first integral,using integration by parts,we get:$$\int_0^{\frac{1}{8}} y \tan(8y)dy = \int_0^{\frac{1}{8}} \frac{y}{8}(-\ln(\cos(8y))'dy=-[\frac{y}{8} \ln(\cos(8y))]_0^{\frac{1}{8}}+ \int_0^{\frac{1}{8}} \frac{1}{8} \ln {\cos(8y)} dy=\int_0^{\frac{1}{8}} \frac{1}{8} \ln {\cos(8y)}dy$$

I think that there isn't a simple way,calculating the integral:
$\int_0^{\frac{1}{8}} \frac{1}{8} \ln {\cos(8y)}dy$

Maybe you could use the formula:

$$\int_0^{\frac{1}{a}} \ln cos(ay)dy = \frac{i}{2a} - \frac{\ln 2}{ a} - \frac{i}{2a} (Li_2(-e^{-2i}) -Li_2(-1)) $$

where Li is the Polylogarithm,see Wikipedia:Polylogarithm - Wikipedia, the free encyclopedia

- - - Updated - - -

For the first integral,using integration by parts,we get:$$\int_0^{\frac{1}{8}} y \tan(8y)dy = \int_0^{\frac{1}{8}} \frac{y}{8}(-\ln(\cos(8y))'dy=-[\frac{y}{8} \ln(\cos(8y))]_0^{\frac{1}{8}}+ \int_0^{\frac{1}{8}} \frac{1}{8} \ln {\cos(8y)} dy=\int_0^{\frac{1}{8}} \frac{1}{8} \ln {\cos(8y)}dy$$

I think that there isn't a simple way,calculating the integral:
$\int_0^{\frac{1}{8}} \frac{1}{8} \ln {\cos(8y)}dy$

Maybe you could use the formula:

$$\int_0^{\frac{1}{a}} \ln cos(ay)dy = \frac{i}{2a} - \frac{\ln 2}{ a} - \frac{i}{2a} (Li_2(-e^{-2i}) -Li_2(-1)) $$

where Li is the Polylogarithm,see Wikipedia:Polylogarithm - Wikipedia, the free encyclopedia

Is there a backround for this exercise? Have you get taught a similar one?

 

[evinda]

ooohhh i get what you did. makes sense. is there any other way to approach this problem or is this the only way? (i have a test the day after tomorrow so i need to REALLY understand this stuff)

I think it is the easiest way to solve this!

 

[ineedhelpnow]

let me check quickly to see if we did anything like that yarctan8y one.

also for the last one was there any way to do the problem without having to multiply 6/6 to the integral?

- - - Updated - - -

Is there a backround for this exercise? Have you get taught a similar one?

i think the closest thing to this problem was an integration by parts problem. the problem is that it has to be a method we used in class otherwise he won't accept it on the test. ill try emailing him about that question though.

he said it needs to be done using IBP. I am not sure what to set as my u though

 

[evinda]

also for the last one was there any way to do the problem without having to multiply 6/6 to the integral?

I think this is the easiest way,because then you have the same expression at the nominator and the denominator,so you can get rid of the $u$,that is at the denominator.And then you have an expression,which antiderivative you know!

he said it needs to be done using IBP. I am not sure what to set as my u though

At the calculation of the last integral ( $ \int \frac{e^{2x}}{6e^x+4}$),we also didn't use the method Integration by parts,but Integration by substitution!

 

[ineedhelpnow]

:) I am familiar with substitutions so he wouldn't mind that but for the first one he said it could be done by IBP. i don't know how to set that up.

 

[Prove It]

$\int \ 6(sec(x))^4,dx$ Reduction

Like I always tell my calculus students, ALWAYS look for a simple substitution first! You should know $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \tan{(x)} \right] = \sec^2{(x)} \end{align*}$ and you should also know $\displaystyle \begin{align*} \sec^2{(x)} \equiv 1 + \tan^2{(x)} \end{align*}$, so

$\displaystyle \begin{align*} \int{ 6\sec^4{(x)}\,\mathrm{d}x } &= 6\int{ \sec^2{(x)}\sec^2{(x)}\,\mathrm{d}x} \\ &= 6\int{ \left[ 1 + \tan^2{(x)} \right] \, \sec^2{(x)}\,\mathrm{d}x} \\ &= 6 \int{ 1 + u^2\,\mathrm{d}u} \textrm{ when we substitute }u = \tan{(x)} \implies \mathrm{d}u = \sec^2{(x)}\,\mathrm{d}x \\ &= 6 \left( u + \frac{u^3}{3} \right) + C \\ &= 6 \left[ \tan{(x)} + \frac{\tan^3{(x)}}{3} \right] + C \end{align*}$

 

[ineedhelpnow]

I was going to do a u-sub but i think my instructor wants us to be able to integrate by reduction on the test

 

[Prove It]

I was going to do a u-sub but i think my instructor wants us to be able to integrate by reduction on the test

Unless you are told to use a specific method, any valid method will be accepted.

 

[ineedhelpnow]

in a way the reduction method was kind of easier since all i had to do was plug the numbers into a already given formula.

 

[evinda]

i think the closest thing to this problem was an integration by parts problem. the problem is that it has to be a method we used in class otherwise he won't accept it on the test. ill try emailing him about that question though.

he said it needs to be done using IBP. I am not sure what to set as my u though

Do you know something more about the first integral? (Worried)

An other way would be to use the expansion of the series$$\tan{x}=\sum_{n=0}^{\infty} \frac{A_{2n+1}}{(2n+1)!}x^{2n+1}$$

where $A_n$ is the n-th alternating permutation.

Then:

$$\int_0^{\frac{1}{8}} y \tan{(8y)}dy=\frac{1}{64} \int_0^1 x \tan{x}dx=\frac{1}{64} \int_0^1 x (\sum_{n=0}^{\infty} \frac{A_{2n+1}}{(2n+1)!}x^{2n+1})dx \\ =\frac{1}{64}\sum_{n=0}^{\infty} \frac{A_{2n+1}}{(2n+1)!} \int_0^1 x^{2n+2}dx=\frac{1}{64} \sum_{n=0}^{\infty} \frac{A_{2n+1}}{(2n+1)!} \frac{1}{2n+3}$$But..then you will have to calculate the series,and it is not sure if we can do this easily. (Thinking)

 

[evinda]

in a way the reduction method was kind of easier since all i had to do was plug the numbers into a already given formula.

You could also do it like that:

$$\int 6 sec^4xdx=6\int \frac{1}{cos^4(x)}dx=6\int \frac{sin^2x+cos^2x}{cos^4x}dx \\ =6\left(\int \frac{sin^2x}{cos^2x}\frac{1}{cos^2x}dx\right)+6\int \frac{1}{cos^2x}dx=6\int tan^2x (tanx)'dx+6tanx=6\frac{1}{3}tan^3x+6tanx+c \\ =2tan^3x+6tanx+c $$

 

[ineedhelpnow]

for the first integral i set my $u=\arctan\left({8y}\right)$ and $dv=ydy$ and then once i used the formula to integrate that i set $w=64y^2+1$ my answer has $ln$ in the final answer but when i put the original equation into my calculator there was no $ln$ and i don't know what I am doing wrong.

 

[I like Serena]

for the first integral i set my $u=\arctan\left({8y}\right)$ and $dv=ydy$ and then once i used the formula to integrate that i set $w=64y^2+1$ my answer has $ln$ in the final answer but when i put the original equation into my calculator there was no $ln$ and i don't know what I am doing wrong.

Hmm... you're mentioning $\arctan$ here...

Do you mean your integral should really be:
$$\int_0^{1/8} y \arctan(8y)\, dy$$
instead of:
$$\int_0^{1/8} y \tan(8y)\, dy$$
(Wondering)

 

[MarkFL]

for the first integral i set my $u=\arctan\left({8y}\right)$ and $dv=ydy$ and then once i used the formula to integrate that i set $w=64y^2+1$ my answer has $ln$ in the final answer but when i put the original equation into my calculator there was no $ln$ and i don't know what I am doing wrong.

Judging from what you've posted here, your "first integral" is actually:

$$I=\int_0^{\frac{1}{8}} y\tan^{-1}(8y)\,dy$$

Now, you did well to let:

$$u=\tan^{-1}(8y)\,\therefore\,du=\frac{8}{64y^2+1}\,dy$$

and so we must have:

$$dv=y\,dy\,\therefore\,v=\frac{1}{2}y^2$$

Hence, we obtain:

$$I=\left[\frac{1}{2}y^2\tan^{-1}(8y)\right]_0^{\frac{1}{8}}-4\int_0^{\frac{1}{8}}\frac{y^2}{64y^2+1}\,dy$$

Now, for the remaining integral, the substitution you tried is not going to work because the numerator has $y^2$ instead of $y$, so I suggest dividing each term by $y^2$ and look at:

$$\int_0^{\frac{1}{8}}\frac{1}{\left(\dfrac{1}{y}\right)^2+8^2}\,dy$$

Now, can you see what substitution would be appropriate here?

 

[ineedhelpnow]

@ilikeserena --- yeah the original equation has arctan not tan

@markfl --- no I am not really sure what to do from there

 

[MarkFL]

@ilikeserena --- yeah the original equation has arctan not tan

@markfl --- no I am not really sure what to do from there

What should you think of when you see the sum of two squares in the denominator of an integrand?

 

[ineedhelpnow]

$\int \,\frac{du}{a^2+u^2}=\frac{1}{a}\arctan\left({\frac{u}{a}}\right) + C$

I think

 

[MarkFL]

$\int \,\frac{du}{a^2+u^2}=\frac{1}{a}\arctan\left({\frac{u}{a}}\right) + C$

I think

Yes, but without relying on this formula, what kind of substitution do we need to make here?

 

[ineedhelpnow]

im not sure

 

[MarkFL]

im not sure

I always think of the Pythagorean identity:

$$\tan^2(\theta)+1=\sec^2(\theta)$$

and so we could look at:

$$\frac{1}{y}=8\tan(\theta)$$

Once we factor out $8^2$ from the denominator, and make the substitution (making sure to change the differential and limits accordingly), what do we have?

 

[ineedhelpnow]

why do you factor out the $8^2$

oh i think i see what you did. i plugged it in for $\frac{1}{y}$ and now i have $\frac{1}{64(sec\theta)^2}$

 

[MarkFL]

why do you factor out the $8^2$

So then we'll have:

$$\tan^2(\theta)+1$$

in the denominator, and we can apply the aforementioned Pythagorean identity...:D