# Evaluating $\int_{1}^{2}\left ( x+3 \right )\ln x \:dx$

I would appreciate some help on evaluating this integral: $$\int_{1}^{2}\left ( x+3 \right )\ln x \:dx$$

I know that I must use integration by parts, but I am stuck at the first step. I chose $u = \ln x$, but what should I do with $(x+3)$, because derivative of that expression is $1$, and it's very strange to me.

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You have:

$$\int_{1}^{2}{(x+3)\ln{x}\:dx}=\int_{1}^{2}{x\ln{x}\:dx+3\int_{1}^{2}{\ln{x}\:dx}}$$

Evaluating the first integral can be done by parts (with $u=\ln{x}$, $v'=x$):

$$\int_{1}^{2}{x\ln{x}\:dx}=\left[\frac{1}{2}x^{2}\ln{x}\right]_{1}^{2}-\frac{1}{2}\int_{1}^{2}{x\:dx}=-\frac{3}{4}+\ln{4}$$

Evaluating the second integral is trivial given the identity: $\int{\ln{x}}\:dx=-x + x \ln{x}+c_{1}$, therefore the second part evaluates to:

$$3\int_{1}^{2}{\ln{x}\:dx}=3\left[-x+x\ln{x}\right]_{1}^{2}=-3+3\ln{4}$$

Combining these gives:

$$\int_{1}^{2}{(x+3)\ln{x}\:dx}=-\frac{3}{4}+\ln{4}-3+3\ln{4}=-\frac{15}{4}+4\ln{4}$$

Hope this helps!

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The identity $$\int{\ln{x}}\:dx=-x + x \ln{x}+c_{1}$$ might not be at the forefront of a beginner's mind. Perhaps pointing out that the antiderivative of $\ln x$ can also be obtained via integration by parts as $$\int{\ln{x}}\:dx=x\ln x - \int{\frac{1}{x}\cdot x}\:dx=x\ln x - x + c$$ might have been more useful. – Dilip Sarwate Sep 3 '12 at 22:07

Try differentiating the log factor and integrating the rest. The $x + 3$ integrates easily to $x^2/2 + 3x$. Don't be afraid to do that.

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I found some interesting method at Wolframalpha link, but i'm also interested in other possible methods. – sundancer Sep 3 '12 at 21:57

Consider $$\int f' \left( x \right) \operatorname{op} \left( g \left( x \right) \right) dx$$ where f, f' and g are rational functions and op is either log, arc-trig or inverse hyperbolic function. Set u = f(x) and v = g(x). Then integrate by parts, resulting in $$\int \operatorname{op} \left( v \right) du = u \operatorname{op} \left( v \right) - \int u \operatorname{op}' \left( v \right) dv.$$

Let op be ln, we obtain $$\int \ln v \, du = u \ln v - \int \frac u v dv.$$

With this technique, we can even integrate $$\int \frac {\ln \left| x^2 + 2x \right|} {x^2 + 2x + 1} dx.$$

For brevity, constants of integration are omitted. It is easy to get $$\int \left( x+3 \right) dx = \frac {x^2} 2 + 3x.$$ Therefore, \begin{align} \int \left( x+3 \right) \ln x \, dx &= \left( \frac{x^2}2 + 3x \right) \ln x - \int \left( \frac x 2 + 3 \right) dx \\ &= \left( \frac{x^2}2 + 3x \right) \ln x - \frac {x^2 + 12x} 4. \end{align}

The rest is easy. $$\int_1^2 \left( x+3 \right) \ln x \, dx = 8 \ln 2 - \frac {15} 4.$$

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