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Math Help - Indefinite integral by compound substitution

  1. #1
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    Indefinite integral by compound substitution

    Need help!

    Integral of (X^5) sqrt (x^3)+1 dx

    I hope this isint too hard to read but its only the sqrt of x^3+1
    Thanks!
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  2. #2
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    \int x^5\sqrt{x^3+1}dx
    try substituting u = x^3+1
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  3. #3
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    Quote Originally Posted by badgerigar View Post
    try substituting u = x^3+1
    I tried that, even split x^5 to (x^2)(x^3) but my answer is way off from the books. btw the answer is 2/15((x^3)+1)^5/2 - 2/9((x^3)+1)^3/2+c
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xclo0sive View Post
    I tried that, even split x^5 to (x^2)(x^3) but my answer is way off from the books. btw the answer is 2/15((x^3)+1)^5/2 - 2/9((x^3)+1)^3/2+c
    Going along with that u-sub, you get \,du=3x^2\,dx

    Thus, you now get \tfrac{1}{3}\int x^3\sqrt{u}\,du

    But u=x^3+1\implies\dots

    So then you should end up with \tfrac{1}{3}\int\dots\,du

    Can you fill in the \dots??
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    Going along with that u-sub, you get \,du=3x^2\,dx

    Thus, you now get \tfrac{1}{3}\int x^3\sqrt{u}\,du

    But u=x^3+1\implies\dots

    So then you should end up with \tfrac{1}{3}\int\dots\,du

    Can you fill in the \dots??
    my final answer was 1/12 (x 2/3((x^3)+1)^3/2+c but its nowhere close to the answer provided.
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  6. #6
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    The book's answer is correct. I dunno what you are doing wrong though.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xclo0sive View Post
    my final answer was 1/12 (x 2/3((x^3)+1)^3/2+c but its nowhere close to the answer provided.
    Quote Originally Posted by Chris L T521 View Post
    Going along with that u-sub, you get \,du=3x^2\,dx

    Thus, you now get \tfrac{1}{3}\int x^3\sqrt{u}\,du

    But u=x^3+1\implies\dots

    So then you should end up with \tfrac{1}{3}\int\dots\,du

    Can you fill in the \dots??
    Since u=x^3+1, it implies that u-1=x^3

    Thus, \tfrac{1}{3}\int x^3\sqrt{u}\,du=\tfrac{1}{3}\int\left(u-1\right)\sqrt{u}\,du=\tfrac{1}{3}\int \left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right)\,du

    Can you take it from here?
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    Going along with that u-sub, you get \,du=3x^2\,dx

    Thus, you now get \tfrac{1}{3}\int x^3\sqrt{u}\,du

    But u=x^3+1\implies\dots

    So then you should end up with \tfrac{1}{3}\int\dots\,du

    Can you fill in the \dots??
    \tfrac{1}{3}\int x^3\sqrt{u}\,du

    u=x^3+1\implies\ x^3 = u - 1


    So \tfrac{1}{3}\int (u - 1)\sqrt{u}\,du = \tfrac{1}{3}\int u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du .

    Can you do this integral?
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  9. #9
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    Quote Originally Posted by Prove It View Post
    \tfrac{1}{3}\int x^3\sqrt{u}\,du

    u=x^3+1\implies\ x^3 = u - 1


    So \tfrac{1}{3}\int (u - 1)\sqrt{u}\,du = \tfrac{1}{3}\int u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du .

    Can you do this integral?
    Oh WOW THanks guys! I completely forgot to substitute the x^3! Thanks!!
    EDIT: How did you get u^ 3/2
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  10. #10
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    How did you get u^ 3/2
    (u-1)\sqrt{u} = u^1u^{1/2}-u^{1/2}
    =u^{1+1/2}-u^{1/2}
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