# Integration by substituition question.

1. Jan 14, 2005

### 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?

2. Jan 14, 2005

### Hurkyl

Staff Emeritus

3. Jan 14, 2005

### dextercioby

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

Daniel.

4. Jan 14, 2005

### MathStudent

5. Jan 14, 2005

### vincentchan

$$\frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2$$
$$= (x^2-1)^1^/^2 (\frac {1}{3}(x^2-1) +1)$$
$$= (x^2-1)^1^/^2 (\frac {1}{3}x^2-\frac {1}{3}+1)$$
$$= (x^2-1)^1^/^2 (\frac {1}{3}x^2+\frac {2}{3})$$
$$\frac {1}{3} (x^2+2)\sqrt{x^2-1}$$ :surprised:

6. Jan 14, 2005

### 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

7. Jan 15, 2005

### misogynisticfeminist

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?

8. Jan 15, 2005

### 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.