Question

$\underline{\underline{\sf Only\:Good\:Users}}$
$\text{Evaluate-}$
$\\\int \dfrac{(x^3+8)(x-1)}{x^2-2x+4} \:dx$
$\begin{gathered} \\\end{gathered} ​$
$\boxed{\text{Topic - Integration\:Class \:12th}}$
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Answer 1

Given :-

$\quad \leadsto \quad \displaystyle \int \sf \dfrac{(x^3+8)(x-1)}{x^2-2x+4} \:dx$

To Find :-

Value of the Integral

Solution :-

Consider $\displaystyle \int \sf \dfrac{(x^3+8)(x-1)}{x^2-2x+4} \:dx$

${ : \implies \quad \displaystyle \int \sf \dfrac{(x³+ 8)(x-1)}{x²-2x+4} \: dx}$

${ : \implies \quad \displaystyle \int \sf \dfrac{(x³+ 2³)(x-1)}{x²-2x+4} \: dx}$

${ : \implies \quad \displaystyle \int \sf \dfrac{(x + 2)(x² - 2 \times \bf x \sf + 2²)(x-1)}{x²-2x+4}\: dx }$

${ : \implies \quad \displaystyle \int \sf \dfrac{(x + 2)(x² - 2x + 4)(x-1)}{x²-2x+4} \: dx }$

${ : \implies \quad \displaystyle \int \sf \dfrac{(x + 2)\cancel{(x² - 2x + 4)}(x-1)}{\cancel{(x²-2x+4)}}\: dx }$

${ : \implies \quad \displaystyle \int \sf \{( x + 2 ) ( x - 1 )\} dx }$

${ : \implies \quad \displaystyle \int \sf \{x ( x - 1 ) + 2 ( x - 1 ) \} \: dx}$

${ : \implies \quad \displaystyle \int \sf ( x² - x + 2x - 2 ) \: \: dx }$

${ : \implies \quad \displaystyle \int \sf ( x² + x - 2 ) dx }$

${ : \implies \quad \displaystyle \int \sf x² dx + \int \sf x \: dx- \int 2 \: dx }$

${ : \implies \quad \displaystyle \sf \dfrac{x^{2+1}}{2+1} + \dfrac{x^{1+1}}{1+1} - 2 \int dx }$

${ : \implies \quad \displaystyle \sf \dfrac{x^3}{3} + \dfrac{x^2}{2} - 2x }$

${ : \implies \quad \displaystyle \bf \dfrac{x^{3}}{3} + \dfrac{x^{2}}{2} - 2x + C }$

$\quad \qquad { \bigstar { \underline { \boxed { \pmb { { \red { \bf { \displaystyle \int \sf \bigg\{ \dfrac{(x^3+8)(x-1)}{x^2-2x+4} \bigg\} \:dx = \displaystyle \bf \dfrac{x^{3}}{3} + \dfrac{x^{2}}{2} - 2x + C }}}}}}}}{\bigstar}$

Used formulae :-

• $\displaystyle \int \sf x^{n} \: dx = \dfrac{x^{n+1}}{n+1} + C$

• $\displaystyle \int \sf dx = x + C$

• ${ \sf a³ + b³ = ( a + b ) ( a² - ab + b²)}$

Additional Information :-

• ${ \boxed { \tt { log ( a ) - log ( b ) = log \bigg ( \dfrac{a}{b} \bigg ) }}}$

• ${ \boxed { \tt log_{a} a = 1 }}$

• ${ \boxed { \tt log_{x} y = \dfrac{log_{e} y}{log_{e} x } }}$

• ${\boxed { \tt Natural \:\: log \:\: of \:\: x = ln ( x ) = log_{e} x }}$

• ${ \boxed { \tt { e = Euler's \:\: number }}}$

• ${ \boxed { \tt { e = \displaystyle \tt \lim_{ \tt n \to \infty } { \bigg ( \tt 1 + \dfrac{1}{n} \bigg )}^{n} }}}$

Maclaurin Series :-

Binomial Expansion For all x² < 1

• ${ \boxed { \tt { \orange { {( 1 \pm x)}^{n} = 1 \pm \dfrac{nx}{1!} + \dfrac{n(n - 1)x²}{2!} + . . . . . . . }}}}$

• ${ \boxed { \tt { \red { {( 1 \pm x)}^{- n} = 1 \mp \dfrac{nx}{1!} + \dfrac{n(n + 1)x²}{2!} + . . . . . . . . }}}}$

Maclaurin Series of Sin x :-

• ${ \boxed { \tt { \orange { Sin x = x - \dfrac{x³}{3!} + \dfrac{x⁵}{5!} - . . . . . . . . . . }}}}$

Maclaurin Series of Cos x :-

• ${ \boxed { \tt { \green { Cos x = 1 - \dfrac{x²}{2!} + \dfrac{x⁴}{4!} - . . . . . . . . . . }}}}$

Maclaurin Series of tan x :-

• ${ \boxed { \tt { \blue { tan x = x + \dfrac{x³}{3} + \dfrac{2x⁵}{15} + . . . . . . . . . . }}}}$

Maclaurin Series of $\bf e^{x}$ :-

• ${ \boxed { \tt { \green { e^{x} = 1 + x + \dfrac{x²}{2!} + \dfrac{x³}{3!} + . . . . . . . . }}}}$

Maclaurin Series of $\bf {tan}^{ - 1 } x$ If & only $\bf { |x| \leqslant 1 }$ :-

• ${ \boxed { \tt { \orange { {tan}^{-1} x = x - \dfrac{x³}{3} + \dfrac{x⁵}{5} - . . . . . . . . . . }}}}$

Commonly Used Derivatives :-

• ${ \boxed { \tt { \red { \dfrac{d}{dx} ( x^{n} ) = n.{(x)^{(n-1)}} }}}}$

• ${ \boxed { \tt { \orange { \dfrac{d}{dx} ( a^{x} ) = a^{x} . log ( a ) }}}}$

• ${ \boxed { \tt { \blue { \dfrac{d}{dx} \bigg ( \dfrac{u}{v} \bigg ) = \dfrac{v . \dfrac{du}{dx} - u . \dfrac{dv}{dx}}{v²} }}}}$

• ${ \boxed { \tt { \green { \dfrac{d}{dx} ( u.v ) = v . \dfrac{du}{dx} + u . \dfrac{dv}{dx} }}}}$