- #1
misogynisticfeminist
- 370
- 0
I'm able to do this question but my answer is different from that in the book.
[tex] \int \frac {x^3}{\sqrt{ x^2 -1}} dx [/tex]
what I did was to take the substituition [tex] u= (x^2-1)^1^/^2 [/tex]
so, [tex] x^2 -1 = u^2 [/tex]
[tex]x^2 = u^2+1 [/tex]
[tex] x= (u^2+1)^1^/^2[/tex]
[tex] x^3= (u^2+1)^3^/^2 [/tex]
[tex] dx= \frac {1}{2}(u^2+1)^-^1^/^2 (2u)du [/tex]
[tex] dx= u(u^2+1)^-^1^/^2 du [/tex]
[tex] \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 [/tex]
which simplifies to,
[tex] \int u^2+1 du [/tex]
[tex] \frac {1}{3} u^3 +u+C [/tex]
[tex] \frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2 [/tex]
that's my final answer but the book gave,
[tex] \frac {1}{3} (x^2+2)\sqrt{x^2-1}+C [/tex]
where does my mistake lie?
[tex] \int \frac {x^3}{\sqrt{ x^2 -1}} dx [/tex]
what I did was to take the substituition [tex] u= (x^2-1)^1^/^2 [/tex]
so, [tex] x^2 -1 = u^2 [/tex]
[tex]x^2 = u^2+1 [/tex]
[tex] x= (u^2+1)^1^/^2[/tex]
[tex] x^3= (u^2+1)^3^/^2 [/tex]
[tex] dx= \frac {1}{2}(u^2+1)^-^1^/^2 (2u)du [/tex]
[tex] dx= u(u^2+1)^-^1^/^2 du [/tex]
[tex] \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 [/tex]
which simplifies to,
[tex] \int u^2+1 du [/tex]
[tex] \frac {1}{3} u^3 +u+C [/tex]
[tex] \frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2 [/tex]
that's my final answer but the book gave,
[tex] \frac {1}{3} (x^2+2)\sqrt{x^2-1}+C [/tex]
where does my mistake lie?