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?

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Answer 1 · 2018-01-28T06:07:33

Please see below.

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

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

= $3-2ln2$

= $3-ln4$