Question

can help me solve the operation and explain how it's done
integrate from x = 3 to 4
e^x / ((e^x) -5) ((e^2x) - 10e^x )^1/2

Answer 1

i can't read it, and i have to run, but i do have a suggestion
make \(e^x\) in to some variable like \(z\)
then it will easier to see what to do
the fact that you have a square root in the denominator with a factor out front suggests that you will probably need a trig sub.,

Answer 2

\[\frac{z}{(z-5)\sqrt{z^2-10z})}\] may be easier to work with

Answer 3

The substitution \(z=e^x\) gives \(dz=e^x~dx\), so the new integral will be
\[\large \int_{e^3}^{e^4}\frac{dz}{(z-5)\sqrt{z^2-10z}}\]
At this point you might want to complete the square under the radical,
\[z^2-10z=z^2-10z+25-25=(z-5)^2-25\]
and so you have
\[\large \int_{e^3}^{e^4}\frac{dz}{(z-5)\sqrt{(z-5)^2-25}}\]
and another substitution simplifies this somewhat, \(u=z-5\) so \(du=dz\).
\[\large \int_{e^3-5}^{e^4-5}\frac{du}{u\sqrt{u^2-25}}\]
which sets you up nicely for a trig sub, \(u=5\sec t\) and \(du=5\sec t\tan t~dt\).
\[\large \int_{\sec^{-1}\left(\frac{e^3-5}{5}\right)}^{\sec^{-1}\left(\frac{e^4-5}{5}\right)}\frac{5\sec t\tan t}{5\sec t\sqrt{(5\sec t)^2-25}}~dt\]