Question

∫((3x-1)/(x^2+1)) dx

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \int \dfrac{3x - 1}{ {x}^{2} + 1 } \: dx$

$\displaystyle = \int \dfrac{3x}{ {x}^{2} + 1 } \: dx - \int \dfrac{1}{ {x}^{2} + 1 } \: dx$

$\displaystyle = \dfrac{3}{2} \int \dfrac{2x}{ {x}^{2} + 1 } \: dx - \int \dfrac{1}{ {x}^{2} + 1 } \: dx$

$\displaystyle = \dfrac{3}{2} \ln \left| {x}^{2} + 1 \right| - \tan^{ - 1} (x) + c$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \: \frac{3x - 1}{ {x}^{2} + 1 } \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \: \frac{3x}{ {x}^{2} + 1} \: dx \: - \: \displaystyle\int\rm \: \frac{1}{ {x}^{2} + 1} \: dx$

$\rm \:  =  \: \dfrac{3}{2} \displaystyle\int\rm \: \frac{2x}{ {x}^{2} + 1} \: dx \: - \: {tan}^{ - 1}x + c$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm \: \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c \: }}}$

Here,

$\red{\rm :\longmapsto\:\boxed{\tt{ ( {x}^{2} + 1)' = 2x \: }}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{3}{2}log | {x}^{2} + 1| - {tan}^{ - 1}x + c$

Hence,

$\boxed{\tt{ \displaystyle\int\rm \: \frac{3x - 1}{ {x}^{2} + 1 } dx=  \: \dfrac{3}{2}log | {x}^{2} + 1| - {tan}^{ - 1}x + c \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$