The formula for integrating by parts is
\\[ \\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \\]
We choose $u = \\simplify[std]{{a}x+{b}}$ and $\\displaystyle\\frac{dv}{dx} = \\simplify[std]{e^({c}x)}$.
\nSo $\\displaystyle \\frac{du}{dx} = \\var{a}$ and $\\displaystyle v = \\simplify[std]{(1/{c})*e^({c}*x)}$.
\nHence,
\\[ \\begin{eqnarray} \\int \\simplify[std]{({a}*x+{b})*e^({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{({1}/{c})*({a}x+{b})*e^({c}x) - (1/{c})*Int(({a})*e^({c}x),x)} \\\\ &=& \\simplify[std]{(1/{c})*({a}x+{b})*e^({c}x) -({a}/{c^2})*e^({c}x) + C}\\\\ &=& \\simplify[std]{(({a}x+{b})/{c}-{a}/{c^2})*e^({c}*x) + C}\\\\ &=& \\simplify[std]{(({a}/{c})x+{b*c-a}/{c^2})*e^({c}*x) + C} \\end{eqnarray} \\]
Hence $\\displaystyle \\simplify[std]{g(x)=({a}/{c})*x+{c*b-a}/{c^2}}$
\nFor this part we choose $u = \\simplify[std]{({a}x+{b})^2}$ and $\\displaystyle \\frac{dv}{dx} = \\simplify[std]{e^({c}x)}$.
\nSo $\\displaystyle \\frac{du}{dx}$ = $\\simplify[std]{{2*a}*({a}*(x)+{b})}$ and $\\displaystyle v = \\simplify[std]{(1/{c})*e^({c}*x)}$.
\nHence,
\\[ \\begin{eqnarray*}I= \\int \\simplify[std]{({a}*x+{b})^2*e^({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{({1}/{c})*({a}x+{b})^2*e^({c}x) - (1/{c})*Int({2*a}*({a}x+{b})*e^({c}x),x)} \\\\ &=& \\simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*Int(({a}x+{b})*e^({c}x),x)}\\dots (*) \\end{eqnarray*}\\]
But in Part a we have aready worked out $\\displaystyle \\simplify[std]{Int(({a}x+{b})*e^({c}*x),x)=(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}$
\nSo on substituting this in equation (*) we find:
\\[ \\begin{eqnarray*}I&=& \\simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}\\\\ &=& \\simplify[std]{({a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3})*e^({c}x) +C} \\end{eqnarray*}\\]
Hence $\\displaystyle \\simplify[std]{h(x)={a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3}}$
\nMore information on the integration of exponentials is in chapter 3 of your Mathematical Physics book, section 3.8 on page 73-75.
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\\[ \\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \\]
Do not input numbers as decimals, only as integers without the decimal point, or fractions
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You are given that the answer is of the form \\[I=g(x)e^{\\var{c}x}+C\\] for a polynomial $g(x)$. You have to find $g(x)$.
$g(x)=\\;$[[0]]
\nInput all numbers as fractions or integers and not decimals.
\nYou can get help by clicking on Show steps. You will not lose any marks if you do.
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\n$\\displaystyle I=\\int \\simplify[std]{({a}x+{b})^2*e^({c}x)} dx $
\nYou are given that the answer is of the form \\[I=h(x)e^{\\var{c}x}+C\\] for a polynomial $h(x)$. You have to find $h(x)$.
\n$h(x)=\\;$[[0]]
\nInput all numbers as fractions or integers and not decimals.
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\nInput all numbers as fractions or integers and not decimals.
\n ", "preamble": {"css": "", "js": ""}, "contributors": [{"name": "Phil Barker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/624/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Phil Barker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/624/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}