적분 전략 고르기

(a) Integrate by parts with $u=\log x$ and $\mathrm{d}v=x\,\mathrm{d}x$, so that $\mathrm{d}u=\frac{\mathrm{d}x}{x}$ and $v=\tfrac{1}{2} x^2$.

$$\begin{aligned} \int x\,\log x\ \mathrm{d}x= \frac{1}{2} x^2\log x- \frac{1}{2} \int x^2\cdot\frac{1}{x}\,\mathrm{d}x =\frac{1}{2} x^2\log x -\frac{1}{4} x^2+C \end{aligned}$$

(b) The denominator is an irreducible quadratic, so partial fractions can't get us any further. To integrate a function whose denominator is quadratic, we split the numerator up so that one piece can be evaluated with a $u$-substitution, and the other piece looks like arctangent.

$$\begin{aligned} \int\frac{(x-1)\mathrm{d}x}{x^2+4x+5} &=\int\frac{x+2-3}{x^2+4x+5}\mathrm{d}x \\ &=\frac{1}{2}\int\frac{2x+4}{x^2+4x+5}\mathrm{d}x -\int\frac{3}{x^2+4x+5}\mathrm{d}x \\ &=\frac{1}{2}\int\frac{2x+4}{x^2+4x+5}\mathrm{d}x-3\int\frac{1}{(x+2)^2+1}\mathrm{d}x \\ &=\frac{1}{2} \log [x^2+4x+5] -3\arctan(x+2)+C \end{aligned}$$

For the last step, you can guess the antiderivative, or use the substitutions $u_1=x^2+4x+5$ and $u_2=x+2$, respectively, for the two integrals.

(c) We use partial fractions.

$$\begin{aligned} \frac{1}{x^2-4x+3} &=\frac{1}{(x-3)(x-1)} =\frac{a}{x-3}+\frac{b}{x-1}\\ 1&=a(x-1)+b(x-3) \end{aligned}$$

Setting $x=3$ gives $a=\frac{1}{2}$. Setting $x=1$ gives $b=-\frac{1}{2}$. So,

$$\begin{aligned} \int\frac{\mathrm{d}x}{x^2-4x+3} =\int\Big(\frac{ 1/2}{x-3} -\frac{ 1/2}{x-1}\Big)\,\mathrm{d}x =\frac{1}{2}\log|x-3|-\frac{1}{2}\log|x-1|+C \end{aligned}$$

(d) Substitute $y=x^3$, $\mathrm{d}y=3x^2\,\mathrm{d}x$.

$$\begin{aligned} \int\frac{x^2\,\mathrm{d}x}{1+x^6}= \frac{1}{3} \int \frac{\mathrm{d}y}{1+y^2} = \frac{1}{3}\arctan y+C =\frac{1}{3}\arctan x^3 +C \end{aligned}$$