Let f(x)= 3x^3+6x-8÷x(x^2+2) (i) Express f(x) in the form (A)+(B÷x)+(Cx+D÷x^2+2). (ii) show the integral 2 to 1 f(x) dx=3-ln4 Can someone please solve this question?

1 Answer
Jan 28, 2018

Please see below.

Explanation:

As #f(x)=(3x^3+6x-8)/(x(x^2+2))=(3x^3+6x-8)/(x^3+2x)#

= #3-8/(x^3+2x)=3-8/(x(x^2+2x)#

Here we have divided #3x^3+6x-8# by #x^2+2x#, which leads to #A=3#

Let #8/(x(x^2+2x))=B/x+(Cx+D)/(x^2+2)#

then #8=B(x^2+2)+x(Cx+D)#

Comparing coeeficients of like powers, we get

#2B=8# i.e. #B=4#
#B+C=0# i.e. #C=-4#
#D=0#

Hence #f(x)=(3x^3+6x-8)/(x(x^2+2))=3-4/x+(4x)/(x^2+2)#

and #I=int_1^2(3x^3+6x-8)/(x(x^2+2))dx#

= #int_1^2(3-4/x+(4x)/(x^2+2))dx#

= #3int_1^2dx-4int_1^2(dx)/x+2int_1^2(2x)/(x^2+2)dx#

= #[3x-4lnx]_1^2+2int_1^2(2x)/(x^2+2)dx#

For #int_1^2(2x)/(x^2+2)dx#, let #u=x^2+2#, then #du=2xdx#

and #int_1^2(2x)/(x^2+2)dx=int_3^6(du)/u=[ln6-ln3]#

Hence #I=[3x-4lnx]_1^2+2[ln6-ln3]#

= #6-4ln2-3+2ln2#

= #3-2ln2#

= #3-ln4#