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Find $\\displaystyle \\int x(a x ^ 2 + b)^{m}\\;dx$
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the following integral using integration by substitution.
\nInput the constant of integration as $C$.
", "advice": "\n\t \n\t \n\tThis exercise is best solved by using substitution.
Note that if we let $u=\\simplify[std]{{a} * (x ^ 2) + {b}}$ then $du=\\simplify[std]{({2*a} * x)*dx }$
Hence we can replace $xdx$ by $\\frac{1}{\\var{2*a}}du$.
Hence the integral becomes:
\n\t \n\t \n\t \n\t\\[\\begin{eqnarray*} I&=&\\simplify[std]{Int((1/{2*a})u^{m},u)}\\\\\n\t \n\t &=&\\simplify[std]{(1/{2*a})u^{m+1}/{m+1}+C}\\\\\n\t \n\t &=& \\simplify[std]{({a} * (x ^ 2) + {b})^{m+1}/{2*a*(m+1)}+C}\n\t \n\t \\end{eqnarray*}\\]
\n\t \n\t \n\t \n\tA Useful Result
This example can be generalised.
Suppose \\[I = \\int\\; f'(x)g(f(x))\\;dx\\]
The using the substitution $u=f(x)$ we find that $du=f'(x)\\;dx$ and so using the same method as above:
\\[I = \\int g(u)\\;du \\]
And if we can find this simpler integral in terms of $u$ we can replace $u$ by $f(x)$ and get the result in terms of $x$.
\\[I=\\simplify[std]{Int( x*({a} * x ^ 2 + {b})^{m},x)}\\]
\n$I=\\;$[[0]]
\nInput numbers in your answer as integers or fractions and not as decimals.
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