How do I find #F(x)=int(x^2+5x+6)/(x+1)dx#?

1 Answer
Monzur R.
Jul 12, 2017

#F(x)=1/2(x^2+8x+4ln|x+1|)+"c"#

Explanation:

Since #F(x)# is usually notation for an indefinite integral, I shall assume the question requires #int(x^2+5x+6)/(x+1)dx#

#int(x^2+5x+6)/(x+1)dx=int((x+1)(x+4)+2)/(x+1)dx#

#=intx+4+2/(x+1)dx=1/2x^2+4x+2ln|x+1|+"c"#

#=1/2(x^2+8x+4ln|x+1|)+"c"#

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