Solve the given problem that involves integration

[chwala]

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.

 

[Mark44]

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.

$7e^x -8 = A(e^x-2)+B$
$A=7$
$-2A+B=-8, ⇒B=6$

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

$u=e^x-2$
$du=e^x dx$
$dx=\dfrac{du}{e^x}$
...
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 $dx = \frac{du}{e^x} = \frac {du}{u + 2}$

$u=e^x-2$
$e^x=u+2$

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

...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]

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.

 

[chwala]

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.

 

[chwala]

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!!