I'm able to do this question but my answer is different from that in the book.(adsbygoogle = window.adsbygoogle || []).push({});

[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?

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# Homework Help: Integration by substituition question.

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