$\int (\ln x)^n dx$ in the most simplified form. Take care to compact long sums into $\sum$ expressions.
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- Find
$\int xe^x dx.$ - Let
$a_n=n^2+3^na_{n-1},$with$a_0=1.$Give a closed form expression for$a_n$using sigma notation.
- Find
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How could you find
$\int (\ln x)\ dx?$ -
How about treating the integrand as
$1\cdot \ln x?$ -
… and using integration by parts.
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Why not apply this same technique to the original integral?
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Try treating it as a recurrence relation.
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… and unravelling it to obtain a
$\sum$expression.
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Let
$I_n=\int (\ln x)^ndx.$We can obtain a recursive expression using integration by parts. We treat the integrand as the product$1 \cdot (\ln x)^n$which gives us$I_n= x \ln^n(x) - \int n\ln^{n-1}(x)\ dx,$and we get the recurrence relation$I_n = x \ln^n(x) - n I_{n-1}.$We can expand out some of the terms to establish the pattern:$$ I_n= x \ln^n(x) - xn\ln^{n-1}(x) + xn(n-1)\ln^{n-2}(x)-\cdots $$With each term, the sign flips, the exponent on the logarithm decreases by
$1,$and each term collects factors decreasing from$n.$Expressing this as a$\Sigma$expression, we get$I_n=\sum_{i=0}^n (-1)^i\frac{n!}{(n-i)!}x\ln^{n-i}(x).$