# Solve the given problem that involves integration

• chwala
In summary, the problem asks for an equation that calculates the derivative of a function with respect to a variable, u. Using partial fractions, i have found that...Using partial fractions (repeated factor), i have found that...Partial fractions is one way to evaluate the derivative of a function with respect to a variable, u. The four lines below are not relevant to the problem.Not "also" -- below is what the problem is asking you to do.In addition to the substitutions above, you should include these:##e^x = u + 2##and ##dx = \frac{du}{e^x} = \frac {du}{u + 2
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
Integration

For part (a),

Using partial fractions (repeated factor), i have...

##7e^x -8 = A(e^x-2)+B##

##A=7##

##-2A+B=-8, ⇒B=6##

$$\int {\frac{7e^x-8}{(e^x-2)^2}}dx=\int \left[{\frac{7}{e^x-2}}+{\frac{6}{(e^x-2)^2}}\right]dx$$

##u=e^x-2##
##du=e^x dx##
##dx=\dfrac{du}{e^x}##

...
also

##u=e^x-2##

##e^x=u+2##

$$\int {\frac{7e^x-8}{(e^x-2)^2}}dx=\int \left[{\frac{7}{e^x-2}}+{\frac{6}{(e^x-2)^2}}\right]=\int \left[{\frac{7u+6}{(e^x-2)^2}}\dfrac{du}{e^x}\right]=\int \left[{\frac{7u+6}{u^2(u+2)}}du\right]$$

...part b later...taking a break.

Last edited:
chwala said:
Homework Statement: See attached.
Relevant Equations: Integration

View attachment 327853

For part (a),

Using partial fractions (repeated factor), i have...
Partial fractions is one way to evaluate the integral, but that's not what the problem is asking you to do. The four lines below are not relevant to the problem.
chwala said:
##7e^x -8 = A(e^x-2)+B##
##A=7##
##-2A+B=-8, ⇒B=6##

$$\int {\frac{7e^x-8}{(e^x-2)^2}}dx=\int \left[{\frac{7}{e^x-2}}+{\frac{6}{(e^x-2)^2}}\right]dx$$

##u=e^x-2##
##du=e^x dx##
##dx=\dfrac{du}{e^x}##
...
also
Not "also" -- below is what the problem is asking you to do.

In addition to the substitutions above, you should include these:
##e^x = u + 2##
and ##dx = \frac{du}{e^x} = \frac {du}{u + 2}##
chwala said:
##u=e^x-2##
##e^x=u+2##

$$\int {\frac{7e^x-8}{(e^x-2)^2}}dx=\int \left[{\frac{7}{e^x-2}}+{\frac{6}{(e^x-2)^2}}\right]=\int \left[{\frac{7u+6}{(e^x-2)^2}}\dfrac{du}{e^x}\right]=\int \left[{\frac{7u+6}{u^2(u+2)}}du\right]$$

...part b later...taking a break.
With the substitutions I added, you can replace everything involving ##e^x## and ##dx## in the starting integral with their equivalents in terms of u and du, in one step.

chwala
Mark44 said:
Partial fractions is one way to evaluate the integral, but that's not what the problem is asking you to do. The four lines below are not relevant to the problem.

Not "also" -- below is what the problem is asking you to do.

In addition to the substitutions above, you should include these:
##e^x = u + 2##
and ##dx = \frac{du}{e^x} = \frac {du}{u + 2}##

With the substitutions I added, you can replace everything involving ##e^x## and ##dx## in the starting integral with their equivalents in terms of u and du, in one step.
I will check this out. Thanks.

Mark44 said:
Partial fractions is one way to evaluate the integral, but that's not what the problem is asking you to do. The four lines below are not relevant to the problem.

Not "also" -- below is what the problem is asking you to do.

In addition to the substitutions above, you should include these:
##e^x = u + 2##
and ##dx = \frac{du}{e^x} = \frac {du}{u + 2}##

With the substitutions I added, you can replace everything involving ##e^x## and ##dx## in the starting integral with their equivalents in terms of u and du, in one step.
True, the first part did not require partial fractions...it was straightforward...need to stop overthinking ...cheers @Mark44

...but part (b) will require that...working on it.

for part (b) i have,

$$\int \dfrac{7u+6}{u^2(u+2)} du = \int\left[\dfrac{-2}{u+2}+ \dfrac{2}{u}+\dfrac{3}{u^2}\right]du$$

...

$$=\left[-2\ln (u+2) + 2 \ln u-\dfrac{3}{u}\right]$$

$$=\left[-2\ln e^x + 2 \ln (e^x-2)-\dfrac{3}{e^x-2}\right]$$

on applying the limits i end up with,

$$=\left[2\ln 4-2\ln 6-\dfrac{3}{4}\right] -\left[2\ln 2-2\ln 4-\dfrac{3}{2}\right]$$

$$=\left[\ln \dfrac{4}{9}-\ln \dfrac{1}{4}-\dfrac{3}{4}+\dfrac{3}{2}\right]$$

$$=\left[\ln \dfrac{16}{9}+\dfrac{3}{4}\right]$$

Bingo!

insight welcome guys!!

Last edited:

• Calculus and Beyond Homework Help
Replies
5
Views
785
• Calculus and Beyond Homework Help
Replies
3
Views
496
• Calculus and Beyond Homework Help
Replies
15
Views
914
• Calculus and Beyond Homework Help
Replies
8
Views
867
• Calculus and Beyond Homework Help
Replies
5
Views
873
• Calculus and Beyond Homework Help
Replies
22
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
615
• Calculus and Beyond Homework Help
Replies
17
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
858
• Calculus and Beyond Homework Help
Replies
12
Views
1K