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Math Help - Indefinite Integral Help

  1. #1
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    Indefinite Integral Help

    Can you help show how to solve these??
    Please kindly show the steps.

    1. (x^2/3) - (x^1/4) / (x^1/2) dx
    2. [x(x+2)] / (x^3 + 3x^2 - 5) dx
    3. (X^3)(x^4 - 4x^2 + 4)^1/2 dx
    4. [(x-1)/(x^5)]^1/2 dx
    5. (x^2 + 3)^1/3 (x^5) dx
    6. (x^3 + 3x^2 - 9) / (x + 3)^1/2 dx

    Thanks in advance!!! :)))
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  2. #2
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    Re: Indefinite Integral Help

    1. Write it as \displaystyle \begin{align*} x^{\frac{2}{3}} - x^{-\frac{1}{4}} \end{align*} and integrate term by term.

    2.
    \displaystyle \begin{align*} \int{\frac{x(x + 2)}{x^3 + 3x^2 - 5}\,dx} &= \int{\frac{x^2 + 2x}{x^3 + 3x^2 - 5}\,dx} \\ &= \frac{1}{3} \int{ \left( \frac{3x^2 + 6x}{x^3 + 3x^2 - 5} \right) dx } \end{align*}

    Now you can make the substitution \displaystyle \begin{align*} u = x^3 + 3x^2 - 5 \implies du = 3x^2 + 6x \, dx \end{align*} and integrate.

    3.
    \displaystyle \begin{align*} \int{\frac{x^3}{\sqrt{x^4 - 4x^2 + 4}}\,dx} &= \int{\frac{x^3}{\sqrt{\left( x^2 - 2 \right)^2 }} \, dx} \\ &= \int{ \frac{x^3}{ x^2 - 2 }\, dx} \\ &= \frac{1}{2} \int{ \frac{2x \cdot x^2 }{x^2 - 2} \, dx } \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = x^2 - 2 \implies du = 2x \, dx \end{align*}.
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  3. #3
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    Re: Indefinite Integral Help

    4.

    \int \sqrt{\frac{x-1}{x^5}}dx=\int \frac{\sqrt{x-1}}{x^{2}\sqrt{x}}dx
    =\int \frac{1}{x^2}\sqrt{\frac{x-1}{x}}dx now take \frac{x-1}{x}=t^2 \implies \frac{dx}{x^2}=2tdt
    now the integral becomes \int 2t^{2}dt = 2\frac{t^3}{3}=\frac{2}{3}(\frac{x-1}{x})^{\frac{3}{2}}

    6.

    \int \frac{x^{3}+3x^{2}-9}{\sqrt{x+3}}dx=\int\frac{x^{2}(x+3)-9}{\sqrt{x+3}}dx
    now we integrate separately
    I_{1}=\int\frac{x^{2}(x+3)}{\sqrt{x+3}}dx=\int x^{2}\sqrt{x+3}dx take x+3=t^2 \implies dx=2tdt & x^2=(t^2-3)^{2}
    I_{1}=\int (t^2-3)^{2}t^{2}dt=2\int (t^{6}-t^{4}+9t^{2})=2[\frac{t^7}{7}+\frac{6t^5}{5}+\frac{9t^3}{3}]
    substitute back the value of t to get the value in x
    I_{2}=\int \frac{9}{\sqrt{x+3}} take x+3=t^2 and I_{2} becomes 18(x+3)^{\frac{1}{2}}
    Add I_{1} and I_{2} to get the result

    5.

    Well I tried for sometime & could not do
    Spoiler:
    I approached it like the first one \int\frac{\sqrt[3]{x^2+3}}{x^5}dx =\int\sqrt[3]{\frac{x^2+3}{x^3}}\frac{dx}{x^4}
    take \frac{x^2+3}{x^3}=t^2 \implies \frac{dx}{x^4}=-\frac{3t^2}{x^2+9} now we can try expressing x in terms of t..which wont work i guess...

    Whats the answer ?, does it even have a elementary anti derivative?
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  4. #4
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    Re: Indefinite Integral Help

    Quote Originally Posted by vanillachyrae View Post
    Can you help show how to solve these??
    Please kindly show the steps.

    1. (x^2/3) - (x^1/4) / (x^1/2) dx
    2. [x(x+2)] / (x^3 + 3x^2 - 5) dx
    3. (X^3)(x^4 - 4x^2 + 4)^1/2 dx
    4. [(x-1)/(x^5)]^1/2 dx
    5. (x^2 + 3)^1/3 (x^5) dx
    6. (x^3 + 3x^2 - 9) / (x + 3)^1/2 dx

    Thanks in advance!!! ))
    5.
    \displaystyle \begin{align*} \int{x^5 \left( x^2 + 3 \right)^{\frac{1}{3}} \, dx} &= \frac{1}{2} \int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\,dx } \end{align*}

    Now let \displaystyle \begin{align*} u = x^2 + 3 \implies du = 2x \, dx \end{align*} and the integral becomes

    \displaystyle \begin{align*} \frac{1}{2}\int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\, dx} &= \frac{1}{2} \int{ \left( u - 3 \right)^2 \, u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ \left( u^2 - 6u + 9 \right) u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ u^{\frac{7}{3}} - 6u^{\frac{4}{3}} + 9u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \left( \frac{3}{10}u^{\frac{10}{3}} - \frac{18}{7}u^{\frac{7}{3}} + \frac{27}{4} u^{\frac{4}{3}} \right) + C \\ &= \frac{3}{20} \left( x^2 + 3 \right)^{\frac{10}{3}} - \frac{9}{7} \left( x^2 + 3 \right)^{\frac{7}{3}} + \frac{27}{8} \left( x^2 + 3 \right)^{\frac{4}{3}} + C \end{align*}
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  5. #5
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    Re: Indefinite Integral Help

    Q No 5 AttachedIndefinite Integral Help-integrate.png
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  6. #6
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    Re: Indefinite Integral Help

    Quote Originally Posted by ibdutt View Post
    Q No 5 AttachedClick image for larger version. 

Name:	Integrate.png 
Views:	4 
Size:	12.8 KB 
ID:	27135
    Is this not exactly what I did in the post directly above?
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  7. #7
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    Re: Indefinite Integral Help

    Hmm..seems like I got the whole question 5 wrong...
    @ vanillachyrae : please learn to write in latex(its not at all hard)
    @ Prove it: You were great and quick as usual
    Thanks from Prove It
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  8. #8
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    Re: Indefinite Integral Help

    Quote Originally Posted by Prove It View Post
    5.
    \displaystyle \begin{align*} \int{x^5 \left( x^2 + 3 \right)^{\frac{1}{3}} \, dx} &= \frac{1}{2} \int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\,dx } \end{align*}

    Now let \displaystyle \begin{align*} u = x^2 + 3 \implies du = 2x \, dx \end{align*} and the integral becomes

    \displaystyle \begin{align*} \frac{1}{2}\int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\, dx} &= \frac{1}{2} \int{ \left( u - 3 \right)^2 \, u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ \left( u^2 - 6u + 9 \right) u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ u^{\frac{7}{3}} - 6u^{\frac{4}{3}} + 9u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \left( \frac{3}{10}u^{\frac{10}{3}} - \frac{18}{7}u^{\frac{7}{3}} + \frac{27}{4} u^{\frac{4}{3}} \right) + C \\ &= \frac{3}{20} \left( x^2 + 3 \right)^{\frac{10}{3}} - \frac{9}{7} \left( x^2 + 3 \right)^{\frac{7}{3}} + \frac{27}{8} \left( x^2 + 3 \right)^{\frac{4}{3}} + C \end{align*}
    Also

    5.\displaystyle \begin{align*} \int{x^5 \left( x^2 + 3 \right)^{\frac{1}{3}} \, dx}  \end{align*}
    now let
    \displaystyle \begin{align*} t = x^2 \implies dt = 2x \, dx \end{align*} and the integral becomes

     \frac{1}{2}\int{ t^2(t+3)^{\frac{1}{3}}dt}

    by parts
    let u=t^2 and dv=(t+3)^{\frac{1}{3}}dt
    and once again with integration by parts
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