How do I integrate #int# #3xln(x^2+5)dx#?

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Answer 1 · 2018-06-07T05:18:49

The answer is #=3/2(x^2+5)ln(3x^2+5)-3/2(x^2+5)+C#

Let #u=x^2+5#, #=>#, #du=2xdx#

The integral is

#I=int3xln(x^2+5)dx=3/2intlnudu#

#intfg'=fg-intf'g#

#f=lnu#, #=>#, #f'=1/u#

#g'=1#, #=>#, #g=u#

Therefore,

#I=3/2(u ln u-int1du)=3/2(u ln u-u)#

#=3/2(x^2+5)ln(3x^2+5)-3/2(x^2+5)+C#