Help with an integration problem.

Full solutions are not allowed.

Try the substitution $u=a^5+ax^4$ as the numerator is a constant multiple of the derivative of the denominator.

As, I mentioned ; I did that already here's what I ended up with.
$ln7 * 1/8a$
Hence,that's why am asking for the solution

I make it $\displaystyle \frac{1}{4a}\ln\left(\frac{17}{2}\right)$

As BobbJo said, posting full solutions is against forum rules. Post your working if you'd like someone to check it.

Here's how I approached it.
Find
let $u=a^5+ax^4$
Therefore, $du=4ax^3 dx$
, $dx$ = $\frac{1}{4ax^3}$ $du$
, $x^3 dx$ =$\frac{1}{4a}$ $du$
We will define the boundaries later, integrating:
$\displaystyle \int\frac{x^3}{u}$ $dx$
Since $x^3 dx$ =$\frac{1}{4a}$ $du$
$\displaystyle \int\frac{1}{u}$ * $\frac{1}{4a}$ $du$
Integrating we get;
$ln|u|$ * $\frac{u}{4a}$

Re-defining the limits as we are dealing with $u$;
Since $u=a^5+ax^4$
Where $x$=a, $u$=$2a^5$
,Where $x$=2a, $u$=$17a^5$
So your definite integral will look like this:$\displaystyle \int_{2a^5}^{17a^5}$
Subtracting "$\mathrm{After \ substantiating \ in \ value \ got \ from \ the\ integration }$" the upper limit from the lower limit will result in $\frac{15a^4}{4}$ * $ln\frac{17}{2}$
However, the answer I got is incorrect , point out the mistake.
Thanks.

$\displaystyle \int\frac{1}{u}$ * $\frac{1}{4a}$ $du$
Integrating we get;
$ln|u|$ * $\frac{u}{4a}$


Thanks.

You seem to have acquired an extra "u".

Should be $\displaystyle\ln |u|\times \frac{1}{4a}$



Aside from going wrong earlier, if you had continued along this path, you'd have ended up with:

$\displaystyle \frac{17a^4}{4}\ln 17 - \frac{2a^4}{4}\ln 2$ which doesn't simplify to what you put.

The integration there is incorrect. $\displaystyle \int \dfrac{\mathrm{d}u}{4au} = \frac{1}{4a}\ln|u|$

I haven't worked through the rest of the question, but does that solve your problem?

I see, Thanks. However, in integrating:
$\displaystyle\int\ \frac{1}{u}*\frac{1}{4a}$ $dx$
Why multypling them then integrating? and why not integrating $\frac{1}{u}$ then integrating $\frac{1}{4a}$ as a constant? to be $\frac{u}{4a}$ As this what I did (How I ended up with the extra $u$)

I see, Thanks. However, in integrating:
$\displaystyle\int\ \frac{1}{u}*\frac{1}{4a}$ $dx$
Why multypling them then integrating? and why not integrating $\frac{1}{u}$ then integrating $\frac{1}{4a}$ as a constant? to be $\frac{u}{4a}$ As this what I did (How I ended up with the extra $u$)

Well, 1/4a is just a constant, you can pull it out of the integral.

$\displaystyle \int \frac{1}{4a}\times \frac{1}{u}\; du = \frac{1}{4a}\int\frac{1}{u}\; du$

If you're going to try treat it as a function of u, then you're trying to integrate a product of two functions, in which case you need to use integration by parts. It's massive overkill, but you might find it useful as an exercise to do it that way.

There is no justfitication for integating the product of two functions one at a time and multiplying them. If there were you could have, for example:

$\displaystyle\int x^2\;dx = \int x \times x \;dx = \frac{x^2}{2}\times\frac{x^2}{2}= \frac{x^4}{4}$ which is clearly WRONG.

When you integrate a function multiplied by a constant, you just integrate the function itself and leave the constant multiple unaffected.

eg $\displaystyle \int 5a \times u \ \mathrm{d}u = 5a \times \frac{1}{2}u^2 = \frac{5}{2}au^2 + k$

where $a$ and $k$ are constants.

This example is no different.

As, I mentioned ; I did that already here's what I ended up with.
$ln7 * 1/8a$
Hence,that's why am asking for the solution

I'm not sure where you mentioned that. I must be blind. I feel bad that I can't read minds.