# 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?

Jan 28, 2018

#### Explanation:

As $f \left(x\right) = \frac{3 {x}^{3} + 6 x - 8}{x \left({x}^{2} + 2\right)} = \frac{3 {x}^{3} + 6 x - 8}{{x}^{3} + 2 x}$

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

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

Let $\frac{8}{x \left({x}^{2} + 2 x\right)} = \frac{B}{x} + \frac{C x + D}{{x}^{2} + 2}$

then $8 = B \left({x}^{2} + 2\right) + x \left(C x + D\right)$

Comparing coeeficients of like powers, we get

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

Hence $f \left(x\right) = \frac{3 {x}^{3} + 6 x - 8}{x \left({x}^{2} + 2\right)} = 3 - \frac{4}{x} + \frac{4 x}{{x}^{2} + 2}$

and $I = {\int}_{1}^{2} \frac{3 {x}^{3} + 6 x - 8}{x \left({x}^{2} + 2\right)} \mathrm{dx}$

= ${\int}_{1}^{2} \left(3 - \frac{4}{x} + \frac{4 x}{{x}^{2} + 2}\right) \mathrm{dx}$

= $3 {\int}_{1}^{2} \mathrm{dx} - 4 {\int}_{1}^{2} \frac{\mathrm{dx}}{x} + 2 {\int}_{1}^{2} \frac{2 x}{{x}^{2} + 2} \mathrm{dx}$

= ${\left[3 x - 4 \ln x\right]}_{1}^{2} + 2 {\int}_{1}^{2} \frac{2 x}{{x}^{2} + 2} \mathrm{dx}$

For ${\int}_{1}^{2} \frac{2 x}{{x}^{2} + 2} \mathrm{dx}$, let $u = {x}^{2} + 2$, then $\mathrm{du} = 2 x \mathrm{dx}$

and ${\int}_{1}^{2} \frac{2 x}{{x}^{2} + 2} \mathrm{dx} = {\int}_{3}^{6} \frac{\mathrm{du}}{u} = \left[\ln 6 - \ln 3\right]$

Hence $I = {\left[3 x - 4 \ln x\right]}_{1}^{2} + 2 \left[\ln 6 - \ln 3\right]$

= $6 - 4 \ln 2 - 3 + 2 \ln 2$

= $3 - 2 \ln 2$

= $3 - \ln 4$