Integration by substituition question.

[misogynisticfeminist]

I'm able to do this question but my answer is different from that in the book.

\int \frac {x^3}{\sqrt{ x^2 -1}} dx

what I did was to take the substituition u= (x^2-1)^1^/^2

so, x^2 -1 = u^2

x^2 = u^2+1
x= (u^2+1)^1^/^2
x^3= (u^2+1)^3^/^2
dx= \frac {1}{2}(u^2+1)^-^1^/^2 (2u)du
dx= u(u^2+1)^-^1^/^2 du

\int \frac {x^3}{\sqrt{ x^2 -1}} dx= \int \frac {(u^2+1)^3^/^2}{u} . \frac {u}{(u^2+1)^1^/^2} du

which simplifies to,

\int u^2+1 du
\frac {1}{3} u^3 +u+C
\frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2

that's my final answer but the book gave,

\frac {1}{3} (x^2+2)\sqrt{x^2-1}+C

where does my mistake lie?

 

[Hurkyl]

Are you sure your answer is not the book's answer?

 

[dextercioby]

Yap,it's the same "animal".It's just the fur is a little shady...

Daniel.

 

[MathStudent]

if your ever unsure of your answer ...you can always take the derivative of your answer and see if you get the integrand.

 

[vincentchan]

your answer is fine... you need one more step to get the text answer
\frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2
= (x^2-1)^1^/^2 (\frac {1}{3}(x^2-1) +1)
= (x^2-1)^1^/^2 (\frac {1}{3}x^2-\frac {1}{3}+1)
= (x^2-1)^1^/^2 (\frac {1}{3}x^2+\frac {2}{3})
\frac {1}{3} (x^2+2)\sqrt{x^2-1} :

 

[Curious3141]

Isn't y^(3/2) = (y)(y^(1/2)) ? Rearrange and simplify what you've got and it should come out the same.

EDIT : NM, Vincent has shown the working

 

[misogynisticfeminist]

if your ever unsure of your answer ...you can always take the derivative of your answer and see if you get the integrand.

hey thanks, that's a very good tip !

I am wondering if the answer which I gave is in a different form from the one in the answer script during an exam, will I be penalised?

 

[dextercioby]

Unless your teacher is a narrow minded s.o.b.,i don't see why.If i were u,on this integral i would have gone for another substitution,using hyperbolic sine and cosine.


Daniel.