Question

x^3/(x+1)^2 integral​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{ {x}^{3} }{(x + 1) ^{2} } dx$

$= \int \frac{ {x}^{3} }{x^{2} + 2x + 1 } dx$

$= \int \frac{ x( {x}^{2} + 2x + 1) - 2( {x}^{2} + 2x + 1) + 3x + 2 }{x^{2} + 2x + 1 } dx$

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

$= \frac{ {x}^{2} }{2} - 2x + 3 \int \frac{(x + 1)dx}{(x + 1)^{2} } - \int \frac{dx}{(x + 1)^{2} }$

$= \frac{ {x}^{2} }{2} - 2x + 3 ln(x + 1) + \frac{1}{x + 1} + c$

Answer 2

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

$2. \: \: \: \boxed{ \sf{ \int \: \dfrac{1}{x} dx = log(x) + c}}$

$3. \: \: \: \boxed{ \sf{ \int \: k \: dx = kx + c}}$

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

$\rm :\longmapsto\: Let \: I \: = \: \int\dfrac{ {x}^{3} }{ {(x + 1)}^{2} } dx$

To evaluate this integral, we use method of Substitution,

$\bf \: Put \: x + 1 = y\bf\implies \:dx \: = \: dy \: and \: x \: = y - 1$

So, given integral can be rewritten as

$\rm :\longmapsto\:I \: = \int\dfrac{ {(y - 1)}^{3} }{ {y}^{2} } dy$

$\rm :\longmapsto\:I \: = \int\dfrac{ {y }^{3} - 3 {y}^{2} + 3y - 1 }{ {y}^{2} } dy$

$\rm :\longmapsto\:I = \int \: \bigg(\dfrac{ {y}^{3} }{ {y}^{2} } - \dfrac{3 {y}^{2} }{ {y}^{2} } + \dfrac{3y}{ {y}^{2} } - \dfrac{1}{ {y}^{2} } \bigg) dy$

$\rm :\longmapsto\:I = \int \: \bigg( y - 3 + \dfrac{3}{y} - {y}^{ - 2} \bigg) dy$

$\rm :\longmapsto\:I = \dfrac{ {y}^{2} }{2} - 3y + 3 log(y) - \dfrac{ {y}^{ - 1} }{ - 1} + c$

$\rm :\longmapsto\:I = \dfrac{ {y}^{2} }{2} - 3y + 3 log(y) + \dfrac{1}{ {y} } + c$

$\rm :\longmapsto\:I = \dfrac{ {(x + 1)}^{2} }{2} - 3(x + 1) + 3 log(x + 1) + \dfrac{1}{ {(x + 1)} } + c$

Additional Information :-

$1. \: \: \: \boxed{ \sf{ \int \: sinx \: dx \: = - cosx + c}}$

$2. \: \: \: \boxed{ \sf{ \int \: cosx \: dx = sinx + c}}$

$3. \: \: \: \boxed{ \sf{ \int \: {sec}^{2} x \: dx \: = \: tanx \: + \: c}}$

$4. \: \: \: \boxed{ \sf{ \int \: {cosec}^{2}x \: dx = - cotx + c}}$

$5. \: \: \: \boxed{ \sf{ \int \: secx \: tanx \: dx = secx + c}}$

$6. \: \: \: \boxed{ \sf{ \int \: cosecx \: cotx \: dx = - cosecx + c}}$