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Integration By Substitution

  1. Aug 2, 2008 #1


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    1. The problem statement, all variables and given/known data:
    using the substitution [itex]u=3x+4[/itex], work out:
    [itex]\int 2x \sqrt{3x+4}[/itex]

    2. The attempt at a solution:
    [itex]\int 2x \sqrt{3x+4}[/itex]
    [itex]u=3x+4 \rightarrow \mathrm{d}x = \frac{\mathrm{d}u}{3}[/itex]
    [itex]\int \left( \frac{u-4}{3} \right)\left(u^{\frac12}\right) \ \frac{\mathrm{d}u}{3}[/itex]
    [itex]\frac19 \int u^{\frac32} - 4u^{\frac12} \ \mathrm{d}u[/itex]
    [itex]=\frac19 \left[ \frac{2u^{\frac52}}{5} - \frac{8u^{\frac32}}{3}\right] + c[/itex]
    [itex]=\frac{2u^{\frac52}}{45} - \frac{8u^{\frac32}}{27}}+c[/itex]
    [itex]=\frac{2(3x+4)^{\frac52}}{45} - \frac{8(3x+4)^{\frac32}}{27} + c[/itex]

    3. The problem that I am encountering:
    I was close to the answer but it is incorrect. The correct answer is:
    [itex]\frac{4(3x+4)^{\frac52}}{45} - \frac{16(3x+4)^{\frac32}}{27} + c[/itex]
    Where have I gone wrong? Thanks in advance.
  2. jcsd
  3. Aug 2, 2008 #2
    You forgot to keep the 2 that was in front of the original problem after you made the substitution.
  4. Aug 2, 2008 #3
    Yeah looks like you did not distrbute the 2 when you set x=(u-4)/3. So change your 4th step to a 2/9 outside the integral.
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