Use the substitution to find the integral of
, so .
Therefore, this integral simplifies to .
Rearranging, we get x = , so this becomes x .
Expanding, this becomes .
As everything is now a power of , we can integrate as follows:
Multiplying out the fractions:
Multiplying out the brackets:
And as , we get:
I've checked this using 'Show Steps' after inputting the integral into WolframAlpha, and it appears to be correct.
However, the answer given in both the textbook and WolframAlpha is:
I would like a hint as to how I would progress from my answer to the answer given.