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Mathematics 14 Online
OpenStudy (anonymous):

Been trying to solve this tricky integral for a while, any help would be appreciated: Integrate (cosx+xsinx)/x(x+cosx)

OpenStudy (anonymous):

I tried using wolfram alpha, it does give a final answer for integral, but does not provide the steps. I tried doing a few substitutions but it doesn't seem to be working out.

hartnn (hartnn):

ok,whats the derivative of (x+cos x) ?

hartnn (hartnn):

@battlefire44

OpenStudy (anonymous):

Sorry back

OpenStudy (anonymous):

Deriv of x + cosx is 1 -sinx, are you suggesting we can do something with this?

hartnn (hartnn):

yup u know this: \[\int\limits_{}^{}f \prime (x)/f(x)=\ln|f(x)|+c \]

OpenStudy (anonymous):

Hmm, can you give a quick example, not sure if i seen that generical formula before, so if we have f(x) = x² then 2x/x² = 2/x and hence integral is 2lnx + c ?

hartnn (hartnn):

yup.

OpenStudy (anonymous):

Does that formula have a specefic name, i could look up a proof for it later, but ill take it as a given for now.

hartnn (hartnn):

u can find that in any complete formula list for integration going ahead with question, i will write numerator as: (x+cos x)-x+x sin x because i have (x+cos x) in denominator ok??

OpenStudy (anonymous):

Yep

hartnn (hartnn):

that will be (x+cos x)-x(1-sin x) ok? now separate the numerator and tell me what u get.

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

Appologize for leaving, accidently closed the site, but we will get 1/x + (1-sinx)/(1+cosx)

OpenStudy (noelgreco):

\[\int\limits x^{2} dx = 2\ln x + c ???\]

OpenStudy (anonymous):

and then we can integrate seperately, and arrive at answer, i got it

hartnn (hartnn):

thats correct,tell me final answer.

OpenStudy (anonymous):

lnx + ln(1+cosx) + c

hartnn (hartnn):

ln |x|-ln|x+cos x|+c

OpenStudy (anonymous):

srry i meant, lnx + ln(x+cosx) + c

hartnn (hartnn):

minus.

OpenStudy (anonymous):

yep, thanks soo much, are you a math student?

OpenStudy (anonymous):

oops, forgot the minus, but i get it

hartnn (hartnn):

welcome :) i am an enginneering student.

OpenStudy (anonymous):

ohhh, great, thanks soo much :)

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