# Integration problems

Hello everyone.
I have been trying to do those 2 exercices for a while now and I can't get it.. We just started doing Integration by parts (is that how you call it in english?)
here are the problems

1) $$\int{ (x^3)(e^{x^2})}$$
and
2) $$\int{ \frac{{(x)(e^x)}} { (x+1)^2}}$$

for the 2nd one I tried
u=e^(x^2) dv= x^3
du= 2xe^(x^2) v= (x^4)/4

From there I just get stuck..
For #2, I dont even know how to start it because there are 3 terms.. Can you do that by parts?
I know these are rookies question, but please explain me
thanks :)

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quasar987
Homework Helper
Gold Member
1) Try your luck with u = x² and dv = xe^x² (you can find v by substitution with x² = w)

quasar987
Homework Helper
Gold Member
2) Set

$$u = (x+1)^2 \ \ \ dv = xe^x dx \Rightarrow du = (2x^2 + 4x +2)dx \ \ \ v = \int(xe^x)dx$$

Let us find v by integrating by parts, using

$$w = x \ \ \ dz = e^x dx \Rightarrow dw = dx \ \ \ z = e^x \Rightarrow v = xe^x - \int e^x dx = xe^x - e^x$$

$$v = xe^x - e^x$$

Continue. You will have to do many other integrations by parts to get to the final answer.

dextercioby
Homework Helper
The last one is tricky.

$$\int\frac{x}{(x+1)^{2}} \ dx=\int\frac{(x+1)-1}{(x+1)^{2}} \ dx=\int \frac{dx}{x+1} -\int \frac{d(x+1)}{(x+1)^{2}} =\ln (x+1)+\frac{1}{x+1}$$

Now

$$I:=\int e^{x}\frac{x}{(x+1)^{2}} \ dx$$

can be part integrated to get

$$I=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)-\int e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right) \ dx=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)$$

$$-\int e^{x} \ln (x+1) \ dx-\int \frac{e^{x}}{x+1} \ dx$$

Use part integration for the first of the last 2 integrals

$$I=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)-\left[e^{x}\ln(x+1)\right]+\int \frac{e^{x}}{x+1} \ dx-\int \frac{e^{x}}{x+1} \ dx =\frac{e^{x}}{x+1} + C$$

Daniel.

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quasar987
Homework Helper
Gold Member
Dexter: What does "Docendo discitur" mean? Who's Seneca?

dextercioby