Question

integrate (x³+4x²-3x-2)/(x+2) dx​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{ {x}^{3} + {4x}^{2} - 3x - 2}{x + 2} \: dx$

Since,

• Degree of numerator = 3

and

• Degree of denominator = 1

It means,

• Degree of numerator > Degree of denominator

So,

We have to divide numerator by denominator using long division.

So,

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} + 2x - 7\:\:}}}\\ {\underline{\sf{x + 2}}}& {\sf{\: {x}^{3} + {4x}^{2} - 3x - 2 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: -  {x}^{3} - 2{x}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  2{x}^{2} - 3x - 2 \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  -  2{x}^{2} - 4x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \:  - 7x - 2  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  7x + 14\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 12\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{ {x}^{3} + {4x}^{2} - 3x - 2}{x + 2} \: dx$

$\rm \: = \: \: \displaystyle\int\sf \bigg( {x}^{2} + 2x - 7 + \dfrac{12}{x + 2} \bigg) \: dx$

$\rm \: = \: \: \dfrac{ {x}^{3} }{3} + {x}^{2} - 7x + 12 log(x + 2) + c$

Additional Information :-

$\green{\boxed{ \bf \:\int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}}$

$\green{\boxed{ \bf \:\int \: kdx =kx + c}}$

$\green{\boxed{ \bf \:\int \: cosx \: dx =sinx + c}}$

$\green{\boxed{ \bf \:\int \: sinx \: dx = - \: cosx + c}}$

$\green{\boxed{ \bf \: \int \: \dfrac{1}{x} dx = log(x) + c}}$

$\green{\boxed{ \bf \: \int \: secx \: tanx \: dx =secx + c}}$

$\green{\boxed{ \bf \: \int \: cosecx \: cotx \: dx = - \: cosecx + c}}$

$\green{\boxed{ \bf \: \int \: {sec}^{2}x \: dx = tanx+ c}}$

$\green{\boxed{ \bf \: \int \: {cosec}^{2}x \: dx = - \: cotx+ c}}$

$\green{\boxed{ \bf \: \int \: cotx \: dx = log(sinx) + c}}$

$\green{\boxed{ \bf \: \int \: tanx \: dx = log(secx) + c}}$