Integration by parts of ln[(1+x^3)^(x^2)]

[Ale_ygt]

Here's what I got so far, starting from the beginning:

\(\displaystyle
\int \ln(1+x^3)^{x^2} dx=
\int x^2\cdot\ln(1+x^3) dx=
\int \ln(1+x^3)\cdot D(\frac{x^3}{3}) dx=
\ln(1+x^3) \cdot \frac{x^3}{3} - \frac{1}{3}\int \frac{3x^2}{1+x^3} \cdot x^3 dx
\)

I think I reached a dead end; I even tried the following:

\(\displaystyle
\ln(1+x^3) \cdot \frac{x^3}{3} - \frac{1}{3}\int D(\ln|1+x^3|) \cdot x^3 dx
\)

but it didn't seem to make things any clearer.
According to the textbook I'm using the result should be \(\displaystyle \frac{x^3+1}{3} \cdot \ln(1+x^3) - \frac{x^3}{3} + c \), and theoretically only integration by parts should be used, since u-substitution hasn't been touched yet.
Any help would be greatly appreciated

 

[Deleted member 4993]

Here's what I got so far, starting from the beginning:

\(\displaystyle
\int \ln(1+x^3)^{x^2} dx=
\int x^2\cdot\ln(1+x^3) dx=
\int \ln(1+x^3)\cdot D(\frac{x^3}{3}) dx=
\ln(1+x^3) \cdot \frac{x^3}{3} - \frac{1}{3}\int \frac{3x^2}{1+x^3} \cdot x^3 dx
\)

I think I reached a dead end; I even tried the following:

\(\displaystyle
\ln(1+x^3) \cdot \frac{x^3}{3} - \frac{1}{3}\int D(\ln|1+x^3|) \cdot x^3 dx
\)

but it didn't seem to make things any clearer.
According to the textbook I'm using the result should be \(\displaystyle \frac{x^3+1}{3} \cdot \ln(1+x^3) - \frac{x^3}{3} + c \), and theoretically only integration by parts should be used, since u-substitution hasn't been touched yet.
Any help would be greatly appreciated

A shorter path would be:

substitute

u = 1 + x^3

du = 3 * x^2 dx

\(\displaystyle
\int \ln(1+x^3)^{x^2} dx=
\int x^2\cdot\ln(1+x^3) dx=
\frac{1}{3}\int \ln(u)du =
\frac{1}{3}[u * ln(u) - u] + C =

\frac{1}{3}[(1 + x^3)*ln(1 + x^3)] - \frac{1}{3}(1 + x^3) + C

\)

 

[Ale_ygt]

A shorter path would be:

substitute

u = 1 + x^3

du = 3 * x^2 dx

\(\displaystyle
\int \ln(1+x^3)^{x^2} dx=
\int x^2\cdot\ln(1+x^3) dx=
\frac{1}{3}\int \ln(u)du =
\frac{1}{3}[u * ln(u) - u] + C =

\frac{1}{3}[(1 + x^3)*ln(1 + x^3)] - \frac{1}{3}(1 + x^3) + C

\)

Hah, it certainly beats the mess I ended up with, thanks

Here's what I got with integration by parts(continuing from my previous post):

\(\displaystyle
\ln(1+x^3) \cdot \frac{x^3}{3} - \int \frac{(1+x^3)-1}{1+x^3} \cdot x^2 dx =
\ln(1+x^3) \cdot \frac{x^3}{3} - \int x^2 dx + \frac{1}{3}\int \frac{3x^2}{1+x^3} dx =
\ln(1+x^3) \cdot \frac{x^3}{3} - \frac{x^3}{3} + \frac{1}{3}\ln|1+x^3| + c
\)

Then, in order to convert \(\displaystyle \ln|1+x^3|\) to \(\displaystyle \ln(1+x^3)\), I reasoned that since \(\displaystyle \ln(1+x^3)\) is already present, x's domain in the whole expression is limited to \(\displaystyle
({-1,+\infty})\), and therefore \(\displaystyle \ln|1+x^3| = \ln(1+x^3)\); then it's trivial to get the final result.

Would that be correct? If not, I'll just take the u-substitution method while discarding the messy integration by parts route and move on.

 

[Deleted member 4993]

Hah, it certainly beats the mess I ended up with, thanks

Here's what I got with integration by parts(continuing from my previous post):

\(\displaystyle
\ln(1+x^3) \cdot \frac{x^3}{3} - \int \frac{(1+x^3)-1}{1+x^3} \cdot x^2 dx =
\ln(1+x^3) \cdot \frac{x^3}{3} - \int x^2 dx + \frac{1}{3}\int \frac{3x^2}{1+x^3} dx =
\ln(1+x^3) \cdot \frac{x^3}{3} - \frac{x^3}{3} + \frac{1}{3}\ln|1+x^3| + c
\)

Then, in order to convert \(\displaystyle \ln|1+x^3|\) to \(\displaystyle \ln(1+x^3)\), I reasoned that since \(\displaystyle \ln(1+x^3)\) is already present, x's domain in the whole expression is limited to \(\displaystyle
({-1,+\infty})\), and therefore \(\displaystyle \ln|1+x^3| = \ln(1+x^3)\); then it's trivial to get the final result.

Would that be correct? If not, I'll just take the u-substitution method while discarding the messy integration by parts route and move on.

Easy way to check would be to differentiate your answer and see if you get back your original expression!

Go at it .....

 

[Ale_ygt]

Turns out it is
Though I was referring more about the correctness of the method/reasoning I used to convert \(\displaystyle \ln|1+x^3|\) to \(\displaystyle \ln(1+x^3)\) in this particular instance rather than the correctness of the solution itself, but I'd say it's correct as well.

 

[tkhunny]

Turns out it is
Though I was referring more about the correctness of the method/reasoning I used to convert \(\displaystyle \ln|1+x^3|\) to \(\displaystyle \ln(1+x^3)\) in this particular instance rather than the correctness of the solution itself, but I'd say it's correct as well.

Why did you feel a need to convert anything. You seem to have invented the absolute values concern. Was it mentioned in the original problem statement? No need to make things harder than they have to be.