How do you evaluate the integral $\int {(1 + x)}^{2} d x$ $int(1+x)^2 dx$ ?

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How do you evaluate the integral $\int {(1 + x)}^{2} d x$ $int(1+x)^2 dx$ ?

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Answer 1 · 2014-08-15T21:06:36

$= \frac{x^{3}}{3} + x^{2} + x + C$ $=x^3/3+x^2+x+C$ , where $C$ $C$ is a constant

Explanation :

$I = \int {(1 + x)}^{2} d x$ $I=int(1+x)^2dx$

It can be solved by two methods,

$(I)$ $(I)$

using Integration by Substitution

let's $1 + x = t$ $1+x=t$ $\Rightarrow$ $=>$ $d x = d t$ $dx=dt$

then, $\int t^{2} d t = \frac{t^{3}}{3} + c$ $intt^2dt=t^3/3+c$

Substituting $t$ $t$ back,

$= \frac{{(1 + x)}^{3}}{3} + c$ $=(1+x)^3/3+c$ , where $c$ $c$ is a constant

$= \frac{1}{3} (x^{3} + 3 x^{2} + 3 x + 1) + c$ $=1/3(x^3+3x^2+3x+1)+c$ , where $c$ $c$ is a constant

$= \frac{x^{3}}{3} + x^{2} + x + \frac{1}{3} + c$ $=x^3/3+x^2+x+1/3+c$ , where $c$ $c$ is a constant

$= \frac{x^{3}}{3} + x^{2} + x + C$ $=x^3/3+x^2+x+C$ , where $C$ $C$ is again a constant

$(I I)$ $(II)$

Expanding ${(1 + x)}^{2} = x^{2} + 2 x + 1$ $(1+x)^2=x^2+2x+1$ , we get

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

$= \frac{x^{3}}{3} + 2 \frac{x^{2}}{2} + x + c$ $=x^3/3+2x^2/2+x+c$ , where $c$ $c$ is a constant

$= \frac{x^{3}}{3} + x^{2} + x + c$ $=x^3/3+x^2+x+c$ , where $c$ $c$ is a constant

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How do you evaluate the integral $\int {(1 + x)}^{2} d x$ $int(1+x)^2 dx$ ?

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