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Let's take this as an example.

$$\int (x^{15}\ln x)dx $$

Is there a way to solve it someway clever? Using integration by parts 15 times would be wearisome...

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More swiftly than WHAT? Integrating by parts just once is enough. –  Michael Hardy Dec 27 '12 at 21:11

7 Answers 7

up vote 4 down vote accepted

Set $\ln(x) = t$. Hence, we get $x = e^t \implies dx = e^t dt$. Hence, \begin{align} I_n & = \int x^n \log(x) dx = \int e^{nt} t e^t dt = \int t e^{(n+1)t} dt = \dfrac1{n+1} \int t d(e^{(n+1)t})\\ & = \dfrac1{n+1} \left( te^{(n+1)t} - \int e^{(n+1)t} dt\right) + c = \dfrac1{n+1} \left(te^{(n+1)t} - \dfrac{e^{(n+1)t}}{n+1} \right)+c\\ & = \dfrac1{n+1} \left(\ln(x) - \dfrac1{n+1} \right)x^{n+1} + c \end{align} Set $n=15$, to get what you want.

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Hint: $$\int x^{15}\ln x dx=\frac1{16}\int(x^{16})^{\prime}\ln x dx$$ Then you only need $1$ integration by parts.

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You can use integration by parts and get a formula for the general case.

Note that, first, $\frac{d}{dx}(x \log{x} - x ) = \log{x}$

Now, for the general case, integrate by parts:

$$\int dx \: x^k \log{x} = x^{k+1} (\log{x} - 1) - k \int dx \: x^k \log{x} + k \int dx \: x^k$$

or,

$$ (k+1) \int dx \: x^k \log{x} = x^{k+1} (\log{x} - 1) + \frac{k}{k+1} x^{k+1}$$

Now just plug in $k=15$ and you are done.

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Intgration by parts is, in this case, not wearisome. Let $du=x^{15}\,dx$, and let $v=\ln x$. (Yes, kind of backwards.)

Then we can take $u=\dfrac{x^{16}}{16}$. Also, $dv=\dfrac{1}{x}\,dx$.

So at the next step we are finding $\displaystyle\int \dfrac{x^{16}}{16}\dfrac{1}{x}\,dx$, that is, $\displaystyle\int\dfrac{x^{15}}{16}\,dx$. This one is immediate.

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As noted, the high $x$ exponent does not complicate things too much as $\ln x$ gives something workable once you take the derivative while integrating it will not eliminate $\ln x$ from your integral. If you want it to look a little less formidable though or already have $\int x\ln xdx$ memorized, proper substitution will clean it up.

$$u=x^8,du=8x^7dx$$

$$\int x^{15}\ln xdx=\frac18\int u\ln u^\frac18du=\frac1{64}\int u\ln udu$$

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Integration by parts is not needed. Note that $\dfrac{d}{dt} x^t = x^t \ln(x)$. Now $\displaystyle\int x^t \ dx = \dfrac{x^{t+1}}{t+1} + c$ so $$\int x^t \ln(x)\ dt = \dfrac{d}{dt} \dfrac{x^{t+1}}{t+1} = \dfrac{(t+1) x^{t+1} \ln(x) - x^{t+1}}{(t+1)^2} + c$$

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Integrating by parts once is enough (but integrating by parts 15 times, as suggested in the original posting, is not enough, as you'll find if you actually do it): \begin{align} \int x^{15} \ln x\,dx & = \int \Big(\ln x\Big)\Big(x^{15} \, dx\Big) \\[10pt] & = \int u\,dv \\[10pt] & = uv - \int v\,du \\[10pt] & = \Big(\ln x\Big) \Big( \frac{x^{16}}{16} \Big) - \int \Big(\frac{x^{16}}{16}\Big) \Big(\frac{dx}{x} \Big) \\[10pt] & = \frac{x^{16}\ln x}{16} - \frac{1}{16} \int x^{15}\,dx \\[10pt] & = \frac{x^{16}\ln x}{16} - \frac{1}{16}\cdot\frac{x^{16}}{16} + C. \end{align}

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