Blog: `int (x^3 + 2x^2 + 3x - 2)/(x^2 + 2x + 2)^2 dx` Evaluate the integral

Given to solve

$\displaystyle\int\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

but first we have to find this

$\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{2}{x}+{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{{x}{\left({{x}}^{{2}}+{2}{x}+{2}\right)}+{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{{x}{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}+\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}+\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

$\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}=\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}+\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

Integrating both sides we get:

$\displaystyle\int\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}=\int\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}{\left.{d}{x}\right.}+\int\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

first let us solve,

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{{x}+{1}-{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{{x}+{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}+\int\frac{{-{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

as we know

$\displaystyle\int\frac{{u}}{{{{u}}^{+}{1}}}{\left.{d}{x}\right.}=\frac{{{\ln{{\left({{u}}^{{2}}+{1}\right)}}}}}{{2}}$

...

Given to solve

$\displaystyle\int\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

but first we have to find this

$\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{2}{x}+{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{{x}{\left({{x}}^{{2}}+{2}{x}+{2}\right)}+{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{{x}{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}+\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

= $\displaystyle\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}+\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

Integrating both sides we get:

first let us solve,

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{{x}+{1}-{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{{x}+{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}+\int\frac{{-{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

as we know

$\displaystyle\int\frac{{u}}{{{{u}}^{+}{1}}}{\left.{d}{x}\right.}=\frac{{{\ln{{\left({{u}}^{{2}}+{1}\right)}}}}}{{2}}$

and $\displaystyle\int\frac{{1}}{{{{u}}^{{2}}+{1}}}{\left.{d}{x}\right.}={{\tan}}^{{-{1}}}{u}$

so ,

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{2}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{{x}+{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}-\int\frac{{{1}}}{{{{\left({x}+{1}\right)}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

= $\displaystyle\frac{{{\ln{{\left({{\left({x}+{1}\right)}}^{{2}}+{1}\right)}}}}}{{2}}-{{\tan}}^{{-{1}}}{\left({x}+{1}\right)}$

and now we have to solve ,

$\displaystyle\int\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

this can be solved by integration by parts so,

$\displaystyle\int{u}{v}'={u}{v}-\int{u}'{v}$

let $\displaystyle{u}={x}-{2}{\quad\text{and}\quad}{v}'=\frac{{1}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}$

=> $\displaystyle{u}'={1}$

so, $\displaystyle{v}=\int\frac{{1}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{1}}{{{\left({{\left({x}+{1}\right)}}^{{2}}+{1}\right)}}^{{2}}}{\left.{d}{x}\right.}$

let $\displaystyle{u}={x}+{1},{d}{u}={\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{1}}{{{\left({{\left({u}\right)}}^{{2}}+{1}\right)}}^{{2}}}{d}{u}$

and now $\displaystyle{u}={\tan{{\left(\theta\right)}}}{s}{o},{d}{u}={\left({{\sec}}^{{2}}{\left(\theta\right)}\right)}{d}\theta$

so ,

=> $\displaystyle\int\frac{{1}}{{{\left({{\left({u}\right)}}^{{2}}+{1}\right)}}^{{2}}}{d}{u}$

=> $\displaystyle\int\frac{{1}}{{{\left({{\left({\tan{{\left(\theta\right)}}}\right)}}^{{2}}+{1}\right)}}^{{2}}}{\left({{\sec}}^{{2}}{\left(\theta\right)}\right)}{d}\theta$

=> $\displaystyle\int\frac{{{{\sec}}^{{2}}{\left(\theta\right)}}}{{{\left({{\left({\tan{{\left(\theta\right)}}}\right)}}^{{2}}+{1}\right)}}^{{2}}}{d}\theta$

=> $\displaystyle\int\frac{{{{\sec}}^{{2}}{\left(\theta\right)}}}{{{\left({{\sec}}^{{2}}{\left(\theta\right)}\right)}}^{{2}}}{d}\theta$

=> $\displaystyle\int{{\cos}}^{{2}}{\left(\theta\right)}{d}\theta$

=> we now that $\displaystyle\int{{\cos}}^{{2}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{2}}\right)}{\left({t}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{t}\right)}}}\right)}\right)}$

so ,

$\displaystyle\int{{\cos}}^{{2}}{\left(\theta\right)}{d}\theta={\left(\frac{{1}}{{2}}\right)}{\left(\theta+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left(\theta\right)}\right)}}}\right)}\right)}$

Now ,

$\displaystyle\int\frac{{1}}{{{\left({{\left({x}+{1}\right)}}^{{2}}+{1}\right)}}^{{2}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}{\left(\theta+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left(\theta\right)}\right)}}}\right)}\right)}$

but $\displaystyle{u}={\tan{{\left(\theta\right)}}}{\quad\text{and}\quad}{x}+{1}={u}$

so ,

$\displaystyle=\int\frac{{1}}{{{\left({{\left({x}+{1}\right)}}^{{2}}+{1}\right)}}^{{2}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}{\left(\theta+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left(\theta\right)}\right)}}}\right)}\right)}={\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}$

now , as per $\displaystyle\int{u}{v}'={u}{v}-\int{u}'{v}$

$\displaystyle\int\frac{{{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle{\left[{\left({x}-{2}\right)}{\left({\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}\right)}\right]}$

$\displaystyle-\int{\left({1}\right)}{\left({\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}\right)}{\left.{d}{x}\right.}$

now let us find the value of

$\displaystyle\int{\left({1}\right)}{\left({\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}\right)}{\left.{d}{x}\right.}$

= $\displaystyle\int{\left({\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}{\left.{d}{x}\right.}+\int{\left({\left(\frac{{1}}{{4}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}\right)}{\left.{d}{x}\right.}$

= > let $\displaystyle{{\tan}}^{{-{{1}}}}{\left({x}+{1}\right)}={u}$

$\displaystyle{\tan{{u}}}={x}+{1}$

=> sec^2(u) = dx

so,

= $\displaystyle\int{\left({\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}+{\left({\left(\frac{{1}}{{4}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}\right)}{\left.{d}{x}\right.}$

= = $\displaystyle\int{\left({\left(\frac{{1}}{{2}}\right)}{\left({u}\right)}+{\left({\left(\frac{{1}}{{4}}\right)}{\left({\sin{{\left({2}{u}\right)}}}\right)}\right)}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

= $\displaystyle\int{\left(\frac{{u}}{{2}}+\frac{{\sin{{\left({2}{u}\right)}}}}{{4}}\right)}{{\sec}}^{{{2}{u}}}{d}{u}$

= $\displaystyle\frac{{1}}{{2}}{\left({u}{\tan{{u}}}+{\ln{{\left({\cos{{u}}}\right)}}}\right)}-\frac{{1}}{{4}}{\ln{{\left({\cos{{2}}}{u}+{1}\right)}}}$

= $\displaystyle\frac{{1}}{{2}}{\left({\left({x}+{1}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}+{\ln{{\left({\cos{{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}}}\right)}}}\right)}-\frac{{1}}{{4}}{\ln{{\left({\cos{{2}}}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}+{1}\right)}}}$

so, now ,

$\displaystyle\int\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle\frac{{{\ln{{\left({{\left({x}+{1}\right)}}^{{2}}+{1}\right)}}}}}{{2}}-{{\tan}}^{{-{1}}}{\left({x}+{1}\right)}+{\left[{\left({x}-{2}\right)}{\left({\left(\frac{{1}}{{2}}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}\right)}}}\right)}\right)}\right)}\right]}$

$\displaystyle-\frac{{1}}{{2}}{\left({\left({x}+{1}\right)}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}+{\ln{{\left({\cos{{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}}}\right)}}}\right)}-\frac{{1}}{{4}}{\ln{{\left({\cos{{2}}}{\left({{\tan}}^{{-{1}}}{\left({x}+{1}\right)}\right)}+{1}\right)}}}$

Blog

$\displaystyle\int\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}{\left.{d}{x}\right.}$ Evaluate the integral

No comments:

Post a Comment

What are the problems with Uganda's government?

Evaluate the integral

No comments:

Post a Comment

What are the problems with Uganda&#39;s government?

$\displaystyle\int\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{3}{x}-{2}}}{{{\left({{x}}^{{2}}+{2}{x}+{2}\right)}}^{{2}}}{\left.{d}{x}\right.}$ Evaluate the integral

What are the problems with Uganda's government?