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Math Help - Need some help with this integral

  1. #1
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    Need some help with this integral

    I'm having trouble evaluating thhis integral

    S = integral sign

    S 4 ( 5^4x+9) dx

    I tried using X^n = (X^n+1 / n+1) + C by taking the constant 4 outside and to the left but the result after applying the above rule made no sense.

    I then tried using substitution where 4x+9 = alpha. So the derivative of alpha is just 4 but cannot reconstruct what this will give. Here's what I ended up with:

    alpha = @= 4x+9
    derivative @ = d@ = 4
    d@ / 4 = dx

    thus,

    S 4 (5^4x+9) dx= (5^4x+9) d@/4 but now I don't know what to do.
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  2. #2
    MHF Contributor harish21's Avatar
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     \int 4(5^{4x}+9) dx = 4 [\int 5^{4x} dx + \int 9 dx]

     \int 5^{4x} dx can be integrated this way:

    let u = 5^4x

    so, ln(u) = ln(5)^{4x} \implies ln(u) = 4x \times ln(5) \implies x = \frac{ln(u)}{4ln(5)}

     \therefore dx = \frac{1}{4 ln(5)} \times \frac{1}{u} \mbox{du}

    substitute u and dx to get:

     \int 5^{4x} dx = \int u \frac{1}{4 ln(5)} \times \frac{1}{u} \mbox{du}

    and integrate to complete ....
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  3. #3
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    The exponent of 5 is 4x+9.

    Could I still use the ln(u) method for this?
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  4. #4
    MHF Contributor harish21's Avatar
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    Yes.
    You can do the same as I have done above.

    Start with u = 5^{4x+9} and follow what I did previously(take log and differentiate).....

    PS: Please make your notations clear. Use LaTEX if possible!
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  5. #5
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    Your "power rule", that the anti-derivative of x^n is \frac{1}{n+1}x^{n+1} requires that x be the base not the exponent.

    The derivative of a^x is ln(a) a^x and the anti-derivative is \frac{a^x}{ln(a)} (which you can derive by taking logarithms as harish21 did).

    To integrate \int 5^{4x+9} dx, let u= 4x+ 9 so that du= 4dx, dx= du/4. The integral becomes \frac{1}{4}\int 5^u du= \frac{5^u}{4ln(5)}+ C= \frac{5^{4x+9}}{4ln(5)}+ C.
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  6. #6
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    An alternative method is to write

    \displaystyle{5^{4x+9} = e^{\ln{(5^{4x+9})}}}

    \displaystyle{=e^{(4x+9)\ln{5}}}

    \displaystyle{= e^{(4\ln{5})x + 9\ln{5}}}.


    So \displaystyle{\int{4\cdot 5^{4x+9}\,dx} = \int{4e^{(4\ln{5})x + 9\ln{5}}\,dx}}

    \displaystyle{= \frac{1}{\ln{5}}\int{(4\ln{5})e^{(4\ln{5})x + 9\ln{5}}\,dx}}.


    Now make the substitution \displaystyle{u = (4\ln{5})x + 9\ln{5}} so that \displaystyle{\frac{du}{dx} = 4\ln{5}} and the integral becomes

    \displaystyle{\frac{1}{\ln{5}}\int{(4\ln{5})e^{(4\  ln{5})x + 9\ln{5}}\,dx} = \frac{1}{\ln{5}}\int{e^u\,\frac{du}{dx}\,dx}}

    \displaystyle{=\frac{1}{\ln{5}}\int{e^u\,du}}

    \displaystyle{= \frac{1}{\ln{5}}\,e^u + C}

    \displaystyle{=\frac{1}{\ln{5}}\,e^{(4\ln{5})x + 9\ln{5}} + C}

    \displaystyle{ = \frac{1}{\ln{5}}\,e^{(4x + 9)\ln{5}} + C}

    \displaystyle{= \frac{1}{\ln{5}}\,e^{\ln{(5^{4x + 9})}} + C}

    \displaystyle{= \frac{5^{4x + 9}}{\ln{5}}} + C.
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  7. #7
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    Yep Thanks a lot guys I had the rules mixed up. Solved it. =) Pretty easy too :P
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