Integration by Substitution (1 Viewer)

[SunnyScience]

Integrate

2/(2x+1)^1/2 dx given u^2=2x +1

and

3x^2 (x^3 -1)^1/2 dx given u = x^3 - 1


ty

 

[Timske]

3x^2 (x^3 -1)^1/2 dx given u = x^3 - 1

du/dx = 3x^2
dx=du/3x^2
3x^2 (u)^1/2 du/3x^2 = (u)^1/2 du
integrate (u)^1/2 du
= [2u^3/2]/3
= [2(x^3-1)^3/2]/3
= 2/3*(x^3-1)^3/2 + c

plz no mistakes

 

[Carrotsticks]

3x^2 (x^3 -1)^1/2 dx given u = x^3 - 1

du/dx = 3x^2
dx=du/3x^2
3x^2 (u)^1/2 du/3x^2 = (u)^1/2 du
integrate (u)^1/2 du
= [2u^3/2]/3
= [2(x^3-1)^3/2]/3
= 2/3*(x^3-1)^3/2 + c

plz no mistakes

Correct, but I don't think it's a good idea to mix du and x.

 

[SunnyScience]

Ty
That's what i got for that question as well - the books answer was wrong :/ (Jones & Couchman B2 25.4 Q3d)

Help with the other one was well? :/ That's the one i really had problems with.

 

[Timske]

2/(2x+1)^1/2

does the answer at the book say 2sqrt(2x + 1) + c ?

 

[zeebobDD]

okay for the last one;
u=x^3-1
therefore
du= 3x^2 dx

thus integral (u)^1/2 du
= 2(u^3/2)/3 then always remember to change u back into x^3-1!
+c

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=u=x^3-1 \\ \frac{du}{dx}=3x^2\\ du=3x^2dx\\ \therefore \int 3x^2\sqrt(x^3-1)=\int \sqrt(u)du\\ =\frac{2(u)^\frac{3}{2}}{3}\\ =\frac{2}{3}(x^3-1)^\frac{3}{2}@plus;C" target="_blank"><img src="http://latex.codecogs.com/gif.latex?u=x^3-1 \\ \frac{du}{dx}=3x^2\\ du=3x^2dx\\ \therefore \int 3x^2\sqrt(x^3-1)=\int \sqrt(u)du\\ =\frac{2(u)^\frac{3}{2}}{3}\\ =\frac{2}{3}(x^3-1)^\frac{3}{2}+C" title="u=x^3-1 \\ \frac{du}{dx}=3x^2\\ du=3x^2dx\\ \therefore \int 3x^2\sqrt(x^3-1)=\int \sqrt(u)du\\ =\frac{2(u)^\frac{3}{2}}{3}\\ =\frac{2}{3}(x^3-1)^\frac{3}{2}+C" /></a>

 

[Carrotsticks]

deswa1, did you know that you could use \sqrt{etc etc} instead of what you used there?

 

[SunnyScience]

2/(2x+1)^1/2

does the answer at the book say 2sqrt(2x + 1) + c ?

It is indeed.

How did you get it?

 

[nightweaver066]

It is indeed.

How did you get it?