# Thread: Evaluating Integrals using the Natural Log function? really lost

1. ## Evaluating Integrals using the Natural Log function? really lost

Bare with me, I just registered so I have no idea how to type any signs ... difficult ones will be written out :o)sorry and thanks.

The problem: The integral from 1-e ((x^2+x+1)/x) dx

So far ... I got u= x^2+x+1 ; du= 2x+1 dx ; -1/2du= xdx <-- i'm not quite sure about this

Integral (1-e) ((x^2+x+1)/x) dx = -1/2 Integral (0-1) (u/du) ... And this is where I get stuck. Help please. Thnks

$\int 1-e^{\frac{x^2+x+1}{x}}dx$

3. $\int_1^e\frac{x^2+x+1}{x}dx =\int_1^e\frac{x^2}{x}+\frac{x}{x}+\frac{1}{x}dx = \int_1^ex+1+\frac{1}{x}dx = \int_1^ex~dx+\int_1^e1~dx+\int_1^e\frac{1}{x}~dx$

4. Originally Posted by seld

$\int 1-e^{\frac{x^2+x+1}{x}}dx$
yes seld it is

5. $\int 1-e^{\frac{x^2+x+1}{x}}dx$

You can certainly distribute the x as pick showed you:

$\int 1-e^{x+1+\frac{1}{x}}dx$

can you take it from there?

6. Originally Posted by seld
$\int 1-e^{\frac{x^2+x+1}{x}}dx$

You can certainly distribute the x as pick showed you:

$\int 1-e^{x+1+\frac{1}{x}}dx$

can you take it from there?
Yeah I started doing it that way which is a lot easier but I got
{x^2/2+x+ln abs(x)} - now I just have to plug in e and 1 into the variable right? My integral looks the way seld wrote it out, I think I miss informed you when I said my integral was the way you asked. sorry.

7. This might be a little above your level, maybe not.

So you have:

$\int 1-e^{x+1+\frac{1}{x}}dx$

You can split off the 1.

$\int 1 dx + \int -e*e^{x + \frac{1}{x}}dx$

so you pull the constant -e out.
$\int 1 dx + -e\int e^{x + \frac{1}{x}}dx$

This is where it starts to gets tricky. So the way I solved it was I used the fact that this is true: $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$

So plugging in the x + 1/x you get:

$e^{x+\frac{1}{x}}=\sum_{k=0}^\infty\frac{(x+\frac{ 1}{x})^k}{k!}$

So then you have:

$\int 1 dx + -e\int\sum_{k=0}^\infty\frac{(x+\frac{1}{x})^k}{k!} dx$

Since the integral of a sum is the sum of the integral you can move that outside of the integral.

$\int 1 dx + -e\sum_{k=0}^\infty\int\frac{(x+\frac{1}{x})^k}{k!} dx$

$\int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\int(x+\frac{1}{x})^ kdx$

Then (x+1/x)^k is hard to work with but we have binomial theorem which states: $(a+b)^n=\sum_{i=0}^n {n \choose i}a^ib^{n-i}$

So you plug x and 1/x into binomial theorem and you have:

$\int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\int\sum_{i=0}^k{k \choose i}x^i\frac{1}{x}^{k-i}dx$

You move the sum out again:

$\int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k\int {k \choose i}x^i\frac{1}{x}^{k-i}dx$

the ${k \choose i}$ is a constant, it's also known as k choose i so you can pull it out.

$\int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k {k \choose i}\int \frac{x^i}{x^{k-i}}dx$

So rewriting the term you get:

$\int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k {k \choose i}\int x^{2i-k}dx$

and lastly power rule:

$x + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k {k \choose i}\frac{x^{2i-k+1}}{2i-k+1} + C$

but that's the solution, it's not . . . pretty though I'm not sure if there's a prettier way to write it.