# Integration troubles

Printable View

• Feb 1st 2010, 10:45 PM
MyHairIsOnFire
Integration troubles
Working on a Calculus I review and two integration problems have stumped me:

1. ∫x√(2x+3)

2. ∫(2x-5)/(2x²+2x+2)

Help would be much appreciated because I don't know where to begin on these and have a strong belief that they will be on my quiz tomorrow.
• Feb 1st 2010, 11:01 PM
Calculus26
1. ∫x√(2x+3)

Let u = 2x + 3

x = (u-3)/2

du = 2dx

1. ∫x√(2x+3) = 1/4 integral[(u-3)sqrt(u)du]

= 1/4 integral[u^(3/2)- 3u^(1/2)]du

2. See attachment
• Feb 1st 2010, 11:10 PM
MyHairIsOnFire
Ah, thank you very much. You're a scholar and a gentleman. It was the substitution that was getting me. Totally forgot that was possible. I guess that's what happens when you take Calc II three years after Calc I.
• Feb 1st 2010, 11:23 PM
Soroban
Hello, MyHairIsOnFire!

Here's how I'd do the first one . . .

Quote:

$\displaystyle 1)\;\;\int x\sqrt{2x+3}\,dx$

Let: $\displaystyle u \:=\:\sqrt{2x+3}\quad\Rightarrow\quad x \:=\:\tfrac{1}{2}(u^2-3) \quad\Rightarrow\quad dx \:=\:u\,du$

Substitute: .$\displaystyle \int \tfrac{1}{2}(u^2-3)\cdot u \cdot u\,du \;=\;\frac{1}{2}\int(u^4 - 3u^2)\,du \;=\;\frac{1}{2}\left(\frac{u^5}{5} - u^3\right) + C$ .$\displaystyle =\;\frac{1}{10}\,u^3\left(u^2-5\right) + C$

Back-substitute: .$\displaystyle \frac{1}{10}\left(\sqrt{2x+3}\right)^3\bigg[(\sqrt{2x+3})^2 - 5\bigg] + C$

. . . . . . . . . . $\displaystyle =\;\frac{1}{10}(2x+3)^{\frac{3}{2}}\bigg[2x + 3 - 5\bigg] + C$

. . . . . . . . . . $\displaystyle =\;\frac{1}{10}(2x+3)^{\frac{3}{2}}\bigg[2x - 2\bigg] + C$

. . . . . . . . . . $\displaystyle =\; \frac{1}{10}(2x+3)^{\frac{3}{2}}\cdot 2(x-1) + C$

. . . . . . . . . . $\displaystyle =\; \frac{1}{5}(2x+3)^{\frac{3}{2}}(x-1) + C$

• Feb 2nd 2010, 06:42 AM
Pulock2009
for the second one u could do something like this:
Numerator=A.d(Denominator)/dx+B where A and B are constants.
once u get the constants u get 2 integrals which can be solved by standard methods. hope u get me.
• Feb 2nd 2010, 02:48 PM
oblixps
for the second integral notice that the derivative of the denominator is 4x+2. if the numerator is 4x+2, it would just become a simple u substitution of form 1/u du. so multiply the numerator by 2 (but you also have to divide by 2 and you can factor that out of the integral). the numerator becomes 4x-10. what you want is 4x+2 so add 12 to 4x-10 to get 4x-10+12 = 4x+2. but you also have to subtract 12 so you would have (4x+2)-12. now split the integral up as (4x+2)/(2x²+2x+2) - 12/(2x²+2x+2). use u substitution on the first integral and complete the square on the denominator of the 2nd integral.