Question #6d541

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Question #6d541

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Answer 1 · 2017-12-15T17:24:12

$e^{x} (x^{2} + 1) + c$ $e^x(x^2+1)+c$

Explanation:

Take the integral

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

Expanding the integrand ${(x + 1)}^{2} = x^{2} + 2 x + 1$ $(x+1)^2=x^2+2x+1$

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

separate

$\int e^{x} x^{2} d x + \int 2 e^{x} x d x + \int e^{x} d x$ $inte^x x^2dx+int2e^x xdx+inte^xdx$

For the integrand $e^{x} x^{2}$ $e^x x^2$ , integrate by parts,
$\int f d g = f g - \int g d f$ $intfdg=fg-intgdf$

$f = x^{2}, d g = e^{x} d x$ $f=x^2, dg=e^xdx$
$d f = 2 x d x$ $df=2xdx$ $g = e^{x}$ $g=e^x$

$e^{x} x^{2} - \int e^{x} 2 x d x$ $e^x x^2-inte^x 2xdx$

The integral of $e^{x} = e^{x} + c$ $e^x=e^x+c$

$\int e^{x} x^{2} d x + \int 2 e^{x} x d x + \int e^{x} d x = e^{x} x^{2} - \int e^{x} 2 x d x + \int 2 e^{x} x d x + e^{x} + c$ $inte^x x^2dx+int2e^x xdx+inte^xdx=e^x x^2-inte^x 2xdx+int2e^x xdx+e^x+c$

$\int e^{x} x^{2} d x + \int 2 e^{x} x d x + \int e^{x} d x = e^{x} x^{2} + e^{x} + c$ $inte^x x^2dx+int2e^x xdx+inte^xdx=e^x x^2+e^x+c$

$\int e^{x} x^{2} d x + \int 2 e^{x} x d x + \int e^{x} d x = e^{x} (x^{2} + 1) + c$ $inte^x x^2dx+int2e^x xdx+inte^xdx=e^x(x^2+1)+c$

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Question #6d541

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