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$\int (x+1)\sqrt{5x-6} \text{dx}$

My working so far:

$u=5x-6$

$du=5dx$

$\int\dfrac{x+1}{5}\sqrt u \ \text{du}$

Substituting x

$\int \dfrac{(u+6)/5)+(5/5)}{5}\cdot\sqrt u \ \text{du}$

$\int (u+11)\sqrt u \ \text{du}$

I've got the same powers right when I integrate that, but I have my coefficients wrong. The answer is:

$\dfrac{2}{75}(5x-6)^{5/2}+\dfrac{22}{75}(5x-6)^{3/2}+c$

Where am I going wrong?

Jim
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  • You just lost a $1/5$. There is one from the $du$, and another one from the substitution in $x+1$. The integration of $u^{3/2}$ gives another. – André Nicolas Jun 05 '14 at 18:54

2 Answers2

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Let the integral in question be \begin{align} I = \int (x+1) \sqrt{5x-6} \ dx \end{align} then making the substitution $u = 5x-6$ the integral becomes \begin{align} I &= \int \left( \frac{u+11}{5} \right) \ \sqrt{u} \ \frac{du}{5} \\ &= \frac{1}{25} \int \left[ u^{3/2} + 11 u^{1/2} \right] du \\ &= \frac{1}{25} \left[ \frac{2}{5} \ u^{5/2} + \frac{22}{3} u^{3/2} \right] \end{align} and upon back substitution the result becomes \begin{align} I = \frac{2}{125} (5x-6)^{5/2} + \frac{22}{75} (5x-6)^{3/2} + c. \end{align}

Leucippus
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1

It looks like you made a simple cancellation error: Your key simplification seems to equate

$${((u+6)/5)+(5/5)\over5}=u+11$$

(minor aside: I added an extra parenthesis to "$(u+6)/5)$"). I'm guessing your thinking here was that the expression on the left hand side simplifies to

$${(u+11)/5\over5}$$

and then that $5$ in the "numerator" cancels with the $5$ in the denominator. The problem is, that top $5$ is actually in the denominator of the numerator, so it more properly just goes down and joins (multiplicatively) the other $5$, making the correct result

$${u+11\over25}$$

One other note: You gave the "answer" with $2/75$ as one of its coefficients; the correct coefficient, as Leucippus found, is $2/125$.

Barry Cipra
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