Question

Integral 2 tanxsec^2x /tan^2x+ 3tanx +2​

Answer 1

Step-by-step explanation:

$i = \int \frac{2 \tan(x) \sec^{2} (x) }{ \tan^{2} (x) + 3 \tan(x) + 2 } dx$

$let \: \: \tan(x) = t = > \sec^{2} (x).dx = dt$

$i= \int \frac{2t}{ {t}^{2} + 3t + 2}dt$

$i= \int \frac{2t + 3 - 3}{ {t}^{2} + 3t + 2} dt$

$i= \int \frac{2t + 3}{ {t}^{2} + 3t + 2 } dt - \int \frac{3}{ {t}^{2} + 3t + 2} dt$

Let i=a-b, where,

$a = \int \frac{2t + 3}{ {t}^{2} + 3t + 2 } dt \: \: and \: \: b = \int \frac{3}{ {t}^{2} + 3t + 2} dt$

Solving a and b:

$a = \int \frac{2t + 3}{ {t}^{2} + 3t + 2} dt$

$let \: \: {t}^{2} + 3t + 2 = u = > (2t + 3)dt = du$

$= > a = \int \frac{du}{u}$

$= > a = ln \: |u| + c= ln \: | {t}^{2} + 3t + 2| + c$

Now,

$b = 3 \int \frac{dt}{ {t}^{2} + 3t + 2}$

$= > b = 3 \int \frac{dt}{ {t}^{2} + 2. \frac{3}{2} t + { (\frac{3}{2}) }^{2} - { (\frac{3}{2}) }^{2} + 2 }$

$= > b = \int \frac{dt}{ {(t + \frac{3}{2} )}^{2} + (2 - \frac{9}{4} ) }$

$= > b = \int \frac{dt}{ {(t + \frac{3}{2}) }^{2} - { (\frac{1}{2} )}^{2} }$

$= > b = \frac{1}{2 \times \frac{1}{2} } ln \: | \frac{ \frac{1}{2} + t + \frac{3}{2} }{ \frac{1}{2} - t - \frac{3}{2} } | + c$

$= > b = ln \: | \frac{t + 2}{ - t - 1} | + c$

$= > b = ln \: | \frac{t + 2}{t + 1} | + c$

$i = ln \: | {t}^{2} + 3t + 2| + ln \: | \frac{t + 2}{t + 1} | + c$

$i = ln \: |(t + 2)(t + 1) \frac{(t + 2)}{(t + 1)} |$

$i = ln \: | {(t + 2)}^{2} | + c$

$i = 2 . ln \: | \tan(x) + 2| + c$