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ineedhelpnow
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Im supposed to use integration by parts for this problem but i understand how to.
$$\int \ \frac{xe^{2x}}{(1+2x)^2},dx$$
$$\int \ \frac{xe^{2x}}{(1+2x)^2},dx$$
ineedhelpnow said:Im supposed to use integration by parts for this problem but i understand how to.
$$\int \ \frac{xe^{2x}}{(1+2x)^2},dx$$
ineedhelpnow said: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
evinda said:For the integral $\int \ \frac{e^{2x}}{6e^x+4}dx$ ,set $e^x=u$.
Then $e^x dx=du$
Therefore:
$$\int \frac{e^{2x}}{8e^x+4}dx=\int \frac{u}{8u+4}du=\int \frac{8u}{8(8u+4)}du=\frac{1}{8} \int \frac{8u+4-4}{8u+4}du=\frac{1}{8} \int (1-\frac{4}{8u+4})du=\frac{1}{8}[u- \frac{1}{2} \ln{|2u+1|}]+c=\frac{1}{8}(e^x-\frac{1}{2} \ln{(2e^x+1)})+c=\frac{e^x}{8}-\frac{1}{16} \ln{(2e^x+1)}+c$$
Yes,in the first step it is actually $\int \frac{u}{6u+4} du$,because:ineedhelpnow said: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)?
ineedhelpnow said: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?
evinda said: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
ineedhelpnow said: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 said:Is there a backround for this exercise? Have you get taught a similar one?
ineedhelpnow said:also for the last one was there any way to do the problem without having to multiply 6/6 to the integral?
ineedhelpnow said:he said it needs to be done using IBP. I am not sure what to set as my u though
ineedhelpnow said:$\int \ 6(sec(x))^4,dx$ Reduction
ineedhelpnow said: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
ineedhelpnow said: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
ineedhelpnow said: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.
ineedhelpnow said: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.
ineedhelpnow said: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.
ineedhelpnow said:@ilikeserena --- yeah the original equation has arctan not tan
@markfl --- no I am not really sure what to do from there
ineedhelpnow said:$\int \,\frac{du}{a^2+u^2}=\frac{1}{a}\arctan\left({\frac{u}{a}}\right) + C$
I think
ineedhelpnow said:im not sure
ineedhelpnow said:why do you factor out the $8^2$
ineedhelpnow said:what do i do once i have $\frac{1}{64(sec\theta)^2}$
Integration by parts is a technique used in calculus to integrate a product of two functions. It is based on the product rule for differentiation, and involves choosing one function to be differentiated and the other to be integrated.
To solve an integral using integration by parts, you must first identify the two functions in the product. Then, choose one function to be differentiated and the other to be integrated. Apply the integration by parts formula: ∫udv = uv - ∫vdu. Repeat this process until the integral can be solved.
The formula for integration by parts is: ∫udv = uv - ∫vdu, where u is the function to be differentiated and dv is the function to be integrated.
There is no specific rule for choosing which function to differentiate and which one to integrate. However, a common method is to choose the function that becomes simpler when differentiated. Another approach is to choose the function that appears in the integral multiple times.
To solve the given integral, we can use the integration by parts formula: ∫udv = uv - ∫vdu. Let u = x and dv = e^2x/(1+2x)^2. Then, du = dx and v = -e^2x/(1+2x). Substituting these into the formula, we get: ∫xe^2x/(1+2x)^2 dx = -xe^2x/(1+2x) + ∫e^2x/(1+2x) dx. This integral can be solved using the substitution method.