Question

Integrate

$\sf \int \: \dfrac{2 {t}^{2} }{ {t}^{2} + t - 2} dt$
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Answer 1

[ Kindly refer attached picture for stepwise solution and answer ]

Additional-Information :

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

Learn more :

d(x^n)/dx = n x^(n-1)

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x

d(e^x)/dx = e^x

d(ln x)/dx = 1/x

Answer 2

EXPLANATION.

$\sf \: \implies \: \int \: \dfrac{2 {t}^{2} }{ {t}^{2} + t - 2} dt \\ \\ \sf \: \implies \: \int \: \frac{2 {t}^{2}dt }{ {t}^{2} + t - 2 } \\ \\ \sf \: \implies \: taking \: 2 \: as \: common \: we \: get \\ \\ \sf \: \implies \: 2 \int \: \frac{ {t}^{2} + t - 2 - t + 2}{ {t}^{2} + t - 2} dt$

$\sf \: \implies \: 2 \int \: \dfrac{ {t}^{2} + t - 2}{ {t}^{2} + t - 2} dt \: \: + \: 2\int \: \dfrac{( - t + 2)}{ {t}^{2} + t - 2} dt \\ \\ \sf \: \implies \: 2 \int \: dt \: + \: 2 \int \: \frac{( - t + 2)}{ {t}^{2} + t - 2} dt \\ \\ \sf \: \implies \: solve \: by \: partial \: fraction \: we \: get \\ \\ \sf \: \implies \: \frac{( - t + 2)}{ {t}^{2} + t - 2} = \frac{a}{(t +2)} + \frac{b}{(t - 1)}$

$\sf \: \implies \: \dfrac{( - t + 2)}{(t + 2)(t - 1)} = \dfrac{a(t - 1) + b(t + 2)}{(t + 2)(t - 1)} \\ \\ \sf \: \implies \: - t + 2 = a(t - 1) + b(t + 2) \\ \\ \sf \: \implies \: put \: t \: = 1 \: we \: get \\ \\ \sf \: \implies \: 1 = 3b \\ \\ \sf \: \implies \: b \: = \frac{1}{3}$

$\sf \: \implies \: put \: t \: = - 2 \: we \: get \\ \\ \sf \: \implies \: 4 = - 3a \\ \\ \sf \: \implies \: a \: = \frac{ - 4}{3} \\ \\ \sf \: \implies \: put \: the \: value \: of \: a \: and \: b \: in \: equation$

$\sf \: \implies \: 2 \int \: dt \: + \int \: \dfrac{ \dfrac{ - 4}{3} dt}{(t + 2)} \: + \: \int \: \dfrac{ \dfrac{1}{3}dt }{(t - 1)} \\ \\ \sf \: \implies \: 2[ t \: - \frac{4}{3} ln(t + 2) + \frac{1}{3} ln(t - 1) ]+ c$