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Input all numbers as integers or fractions and not as decimals.

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$I=\\simplify[std]{Int( x*({a[0]}x^2+{b[0]})^{m[0]},x)}$

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Use $u=\\simplify[std]{{a[0]}x^2+{b[0]}}$ as your substitution.

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$\\frac{du}{dx}=$ [[1]]

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$dx=$ [[2]]

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Substituting back into the original equation for $dx$ and pulling out constants gives

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$I=$[[3]]$\\simplify[std]{Int(u^{m[0]},u)}$

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The next step is to integrate.

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$\\simplify{Int(u^{m[0]},u)}=$ [[4]]

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Putting all of these results together, we get the final answer of:

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[[0]]

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Integrate the following by substitution.

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Don't forget the constant of integration ($C$).

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This problem is best solved by using substitution.
Note that if we let $u=\\simplify[std]{{a[0]} * (x ^ 2) + {b[0]}}$ then $du=\\simplify[std]{(2*{a[0]} * x)*dx }$
Hence we can replace $xdx$ by $\\frac{1}{2*\\var{a[0]}}du$.

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Hence the integral becomes:

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\\[\\begin{eqnarray*} I&=&\\simplify[std]{Int((1/(2{a[0]}))u^{m[0]},u)}\\\\ &=&\\simplify[all]{(1/(2{a[0]}))u^{m[0]+1}/{m[0]+1}+C}\\\\ &=& \\simplify[all]{({a[0]} * (x ^ 2) + {b[0]})^{m[0]+1}/(2{a[0]}*({m[0]}+1))+C} \\end{eqnarray*}\\]

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A 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 $x$.

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Step by step solving for integration by substitution

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