Question

can anyone show me the step by step how to solve this ...
\[\int\limits_{}^{} \sinh ^{-1}x dx\]

Answer 1

does inverse of sinh have a nice e format?

Answer 2

\[\sinh ^{-1}x =\ln (x+\sqrt{x^{2}+1})\]

Answer 3

hmmm, yeah

isnt there some rule similar to sin and cos ??

Answer 4

i just havent played with the hypers enough to recall it that well; but this might be useful

http://www.wolframalpha.com/input/?i=integrate+sinh%5E-1+x

Answer 5

tq...I think i have to use integration by part... ;)

Answer 6

integration by parts eh, i was considering something like this

\[y =\ln (x+\sqrt{x^{2}+1})\]
\[\large e^y =e^{ln (x+\sqrt{x^{2}+1})}\]
\[\large e^y =x+\sqrt{x^{2}+1}\]
\[\large\int( e^y =x+\sqrt{x^{2}+1})\]
\[\large e^y =\int x~dx+\int~\sqrt{x^{2}+1}~dx\]
\[\large e^y =\frac12x^2+c+\int~(x^{2}+1)^{1/2}~dx\]
u = x^2+1; du = 2x dx
\[\large e^y =\frac12x^2+c+\int~\frac22(u)^{1/2}~du\]
\[\large e^y =\frac12x^2+c+\frac12\int~2(u)^{1/2}~du\]
\[\large e^y =\frac12x^2+c+\frac12\frac23~2(u)^{3/2}+c\]
\[\large e^y =\frac12x^2+\frac23(u)^{3/2}+C\]
\[\large e^y =\frac12x^2+\frac23(x^2+1)^{3/2}+C\]

Answer 7

i believe i mismathed it at my attempt to usub

Answer 8

oh well :) good luck with it tho, ive got to get ;)

Answer 9

why dont we use...\[u = \ln (x+\sqrt{x^{2}+1})\]
\[dv = dx\]