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Math Help - Indefinite Integrals...I've got a good start but...

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    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?
    Attached Thumbnails Attached Thumbnails Indefinite Integrals...I've got a good start but...-picture-2.png  
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    1. Try u=e^(7x)+3
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    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?
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    Quote Originally Posted by fattydq View Post
    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
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    so it would just be ln(u)+C?
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    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?
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    Quote Originally Posted by fattydq View Post
    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

    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.

    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]
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    Quote Originally Posted by Reckoner View Post
    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
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    Quote Originally Posted by fattydq View Post
    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.
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    Quote Originally Posted by Jameson View Post
    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.
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    Quote Originally Posted by fattydq View Post
    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.
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