# Indefinite Integrals...I've got a good start but...

• Feb 16th 2009, 11:36 AM
fattydq
Indefinite Integrals...I've got a good start but...
A picture is attached...both involve substitution but what I don't understand is for the first one I know u would be e^7x and du would then be e^7x but where does that leave me? How do I decide what becomes u and what remains e^7x in the function?

For the second I know u would be x^3 and du would be 3x^2 so you'd have 1/3 the integral of u times the square root of...SOMETHING. Now would the x^2 inside the sq root just disappear leaving with 42 or what?
• Feb 16th 2009, 11:38 AM
Jameson
1. Try u=e^(7x)+3
• Feb 16th 2009, 11:45 AM
fattydq
Alright so then my next step would be the integral of 1/u and the antiderivative of that would be...what? Because 1/u is just u^-1 so wouldn't the antiderivative just be u^0 or 1?
• Feb 16th 2009, 11:47 AM
Jameson
Quote:

Originally Posted by fattydq
Alright so then my next step would be the integral of 1/u and the antiderivative of that would be...what? Because 1/u is just u^-1 so wouldn't the antiderivative just be u^0 or 1?

$\int \frac{1}{x}dx = \ln(x)+C$
• Feb 16th 2009, 12:06 PM
fattydq
so it would just be ln(u)+C?
• Feb 16th 2009, 12:08 PM
fattydq
It doesn't seem to accept that answer...am I missing something? Because all I was left with was 1/u and if the antiderivative of that is ln(u) then shouldn't that be my solution?
• Feb 16th 2009, 12:44 PM
Reckoner
Quote:

Originally Posted by fattydq
A picture is attached...both involve substitution but what I don't understand is for the first one I know u would be e^7x and du would then be e^7x but where does that leave me? How do I decide what becomes u and what remains e^7x in the function?

Don't forget to use the chain rule: $du = 7e^{7x}\,dx$

Quote:

Alright so then my next step would be the integral of 1/u and the antiderivative of that would be...what? Because 1/u is just u^-1 so wouldn't the antiderivative just be u^0 or 1?
The power rule for integration is $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C, n\neq-1\text.$ Obviously, this cannot work for an exponent of -1, because then there would be a division by zero. Instead, you have to use the log rule, which Jameson listed above.

Quote:

For the second I know u would be x^3 and du would be 3x^2 so you'd have 1/3 the integral of u times the square root of...SOMETHING. Now would the x^2 inside the sq root just disappear leaving with 42 or what?
Let $u = x^2 + 42\text.$ Then $du = 2x\,dx\Rightarrow dx=\frac{du}{2x}\text.$

Substitution gives

$\int x^3\sqrt{x^2+42}\,dx=\int x^3\sqrt u\left(\frac{du}{2x}\right)$

$=\frac12\int x^2\sqrt u\,du$

But here's the kicker: $u=x^2+42$ means that $x^2 = u-42\text.$ So we can substitute for $x^2$ to get

$\frac12\int(u-42)\sqrt u\,du$

$=\frac12\left[\int u\sqrt u\,du - 42\int\sqrt u\,du\right]$

$=\frac12\left[\int u^{3/2}\,du - 42\int u^{1/2}\,du\right]$
• Feb 16th 2009, 01:33 PM
fattydq
Quote:

Originally Posted by Reckoner
Don't forget to use the chain rule: $du = 7e^{7x}\,dx$

The power rule for integration is $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C, n\neq-1\text.$ Obviously, this cannot work for an exponent of -1, because then there would be a division by zero. Instead, you have to use the log rule, which Jameson listed above.

Let $u = x^2 + 42\text.$ Then $du = 2x\,dx\Rightarrow dx=\frac{du}{2x}\text.$

Substitution gives

$\int x^3\sqrt{x^2+42}\,dx=\int x^3\sqrt u\left(\frac{du}{2x}\right)$

$=\frac12\int x^2\sqrt u\,du$

But here's the kicker: $u=x^2+42$ means that $x^2 = u-42\text.$ So we can substitute for $x^2$ to get

$\frac12\int(u-42)\sqrt u\,du$

$=\frac12\left[\int u\sqrt u\,du - 42\int\sqrt u\,du\right]$

$=\frac12\left[\int u^{3/2}\,du - 42\int u^{1/2}\,du\right]$

OK but then I tried taking the antiderivatives of those and failed since it's still wrong :(
• Feb 16th 2009, 01:35 PM
Jameson
Quote:

Originally Posted by fattydq
OK but then I tried taking the antiderivatives of those and failed since it's still wrong :(

Don't just tell us you got them wrong, SHOW us what you did. This is a two way process.
• Feb 16th 2009, 02:35 PM
fattydq
Quote:

Originally Posted by Jameson
Don't just tell us you got them wrong, SHOW us what you did. This is a two way process.

OK so going from what Reckoner said I did

1/2((5u^5/2 divided by 2)-3u^3/2 divided by 2)) and the answer is incorrect. I put in x^2+42 for every u you see in the problem and my final (incorrect) response, did it 5 times with a calculator and got the same answer so I must be doing something wrong with these antiderivatives.
• Feb 16th 2009, 03:21 PM
Reckoner
Quote:

Originally Posted by fattydq
OK so going from what Reckoner said I did

1/2((5u^5/2 divided by 2)-3u^3/2 divided by 2)) and the answer is incorrect. I put in x^2+42 for every u you see in the problem and my final (incorrect) response, did it 5 times with a calculator and got the same answer so I must be doing something wrong with these antiderivatives.

The rule is $\int u^n\,du = \frac{u^{n+1}}{n+1}+C,\;n\neq0\text.$ It looks like you are doing $(n+1)u^{n+1}+C,$ which is incorrect.

You should get:

$\frac12\left[\int u^{3/2}\,du - 42\int u^{1/2}\,du\right]$

$=\frac12\left(\frac{u^{5/2}}{5/2} - \frac{42u^{3/2}}{3/2}\right)+C$

$=\frac12\left(\frac{2u^{5/2}}5 - \frac{84u^{3/2}}3\right)+C$

$=\frac15u^{5/2}-14u^{3/2}+C$

Now back-substitute.