Question

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?

Answer 1

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`