적분 기법 전체 문제

We can use partial fraction decomposition to break this into chunks that we can deal with. The denominator has a repeated linear factor, so it can be decomposed as the sum of constants divided by powers of that factor.

$$\begin{aligned} \frac{3x^2+4x+6}{(x+1)^3} &=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}\\ &=\frac{A(x+1)^2+B(x+1)+C}{(x+1)^3}\\ \Rightarrow ~~~~ 3x^2+4x+6&=A(x+1)^2+B(x+1)+C\\ &=Ax^2+(2A+B)x+(A+B+C) \\ \text{So, by matching coefficients:} \\ &A=3, ~2A+B=4,\text{ and } A+B+C=6\\ &A=3, ~B=-2,~C=5 \\ \text{Therefore:} \\ \frac{3x^2+4x+6}{(x+1)^3} &= \frac{ 3}{x+1}+\frac{ -2}{(x+1)^2}+\frac{ 5}{(x+1)^3} \\ \text{ Now, the integration is easy, with a substitution of $u=x+1$ and $\mathrm{d}u=\mathrm{d}x$:} \\ \int\frac{3x^2+4x+6}{(x+1)^3} \mathrm{d}x&= \int\left(\frac{3}{x+1}+\frac{-2}{(x+1)^2}+\frac{5}{(x+1)^3}\right)\mathrm{d}x\\ &=\int \left( 3u^{-1}-2u^{-2}+5u^{-3}\right)\mathrm{d}u\\ &=3\log|u|+2u^{-1}-\frac{5}{2} u^{-2}+C\\ &=3\log|x+1|+\frac{2}{x+1}-\frac{5}{2(x+1)^2}+C \end{aligned}$$

If the denominator were $x^2+1$, the antiderivative would be arctangent. So, by completing the square, let's aim for the fraction to look like $\dfrac{1}{u^2+1}$, for some $u$. This is a good strategy for integrating an irreducible quadratic under a constant.

$$\begin{aligned} \\ \text{First: complete the square} \\ \int\frac{1}{x^2+x+1}\mathrm{d}x&= \int\frac{1}{x^2+x+\frac{1}{4}+\frac{3}{4}}\mathrm{d}x = \int\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\mathrm{d}x \\ \text{Second: get the denominator in the form $u^2+1$. To do this, we need to fix the constant} \\ &=\int\left(\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)\left(\frac{\frac{4}{3}}{\frac{4}{3}}\right)\mathrm{d}x\\ &=\frac{4}{3}\int\frac{1}{\frac{4}{3}\cdot\left(x+\frac{1}{2}\right)^2+1}\mathrm{d}x \\ \text{Now a quick wiggle to make that first part of the denominator into something squared again:} \\ &=\frac{4}{3}\int\frac{1}{\left(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\right)^2+1}\mathrm{d}x \\ \text{Now we see that $u=\dfrac{2}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}$, $\mathrm{d}u=\dfrac{2}{\sqrt3}\mathrm{d}x$ will do the job} \\ &=\frac{4}{3}\int\frac{1}{u^2+1}\cdot\frac{\sqrt{3}}{2}\mathrm{d}u =\frac{2}{\sqrt3}\int\frac{1}{u^2+1}\mathrm{d}u\\ &=\frac{2}{\sqrt3}\arctan u +C\\ &=\frac{2}{\sqrt3}\arctan\left(\frac{2}{\sqrt3}x+\frac{1}{\sqrt3}\right) +C \end{aligned}$$

Since $\tan x =\frac{\sin x}{\cos x}$,

$$\begin{aligned} \int \sin x \cos x \tan x~\mathrm{d}x&=\int \sin^2 x~\mathrm{d}x = \int \frac{1}{2}\big(1-\cos(2x) \big)~\mathrm{d}x\\ &=\frac12\left(x - \frac{1}{2}\sin(2x) \right)+C \\&=\frac12\left(x -\sin x \cos x \right)+C \end{aligned}$$

We would like to not have that square root there. Luckily, there's a way of turning cosine into cosine squared: the identity $\cos(2x) = 2\cos^2x -1$. If we take $2x=t$, then $\cos t = 2\cos^2(t/2)-1$.

$$\begin{aligned} \int_0^{\pi/2}\sqrt{\cos t+1}\ \mathrm{d}t&=\int_0^{\pi/2}\sqrt{2\cos^2 (t/2)}~\mathrm{d}t = \sqrt{2}\int_0^{\pi/2} |\cos (t/2)|~\mathrm{d}t \\ \text{Over the interval $[0,\frac{\pi}{2}]$, $\cos (t/2) >0$, so we can drop the absolute values.} \\ &= \sqrt{2}\int_0^{\pi/2} \cos (t/2)~\mathrm{d}t = \sqrt{2}\left[2\sin\left(\frac{t}{2}\right)\right]_0^{\pi/2}\\ &=2\sqrt{2}\sin\left(\frac{\pi}{4}\right) = {2} \end{aligned}$$

• Solution 1: Using logarithm rules, $\log\sqrt{x} = \log\left(x^{1/2}\right)=\frac{1}{2}\log x$, so we can simplify: $$\int_1^e \frac{\log\sqrt{x}}{{x}}\mathrm{d}x =\int_1^e \frac{\log{x}}{{2x}}\mathrm{d}x$$

We use the substitution $u=\log x$, $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$:

$$\begin{aligned} \int_1^e \frac{\log{x}}{{2x}}\mathrm{d}x&=\frac{1}{2}\int_1^e \underbrace{\log(x)}_{u}\cdot\underbrace{\frac{1}{x}\,\mathrm{d}x}_{\mathrm{d}u}\\ &=\frac{1}{2}\int_{\log(1)}^{\log(e)}u\,\mathrm{d}u\\ &=\frac{1}{2}\int_{0}^{1}u\,\mathrm{d}u\\ &=\frac{1}{2}\left[ \frac{1}{2}u^2\right]_0^1\\ &=\frac{1}{2}\left[\frac{1}{2}-0\right]=\dfrac{1}{4} \end{aligned}$$

• Solution 2: We use the substitution $u=\log\sqrt{x}$. Then $\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{\sqrt{x}}\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{1}{2x}$, hence $2\mathrm{d}u=\dfrac{1}{x}\mathrm{d}x$. This fits our integral nicely!

$$\begin{aligned} \int_1^e\frac{\log\sqrt{x}}{x}\mathrm{d}x &= \int_{\log\sqrt{1}}^{\log\sqrt{e}}u\cdot 2\mathrm{d}u \\ &=\Big[u^2\Big]_{0}^{1/2}\\ &=\left(\frac{1}{2}\right)^2-0^2=\frac{1}{4} \end{aligned}$$

$$\int_{0.1}^{0.2} \frac{\tan x}{\log(\cos x)} \mathrm{d}x$$

It might not be immediately obvious how to proceed on this one, so this is another example of an integral where you should not be discouraged by finding methods that don't work. One thing that's worked for us in the past is to use a $u$-substitution with the denominator. With that in mind, let's find the derivative of the denominator. $$\frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(\cos x)\right\}= \frac{1}{\cos x}\cdot (-\sin x) = \frac{-\sin x}{\cos x} = -\tan x$$

So, if we let $u=\log(\cos x$), we see $-\mathrm{d}u=\tan x \mathrm{d}x$, which will work for a substitution.

$$\begin{aligned} \int_{0.1}^{0.2} \frac{\tan x}{\log(\cos x)} \mathrm{d}x&= \int_{\log(\cos(0.1))}^{\log(\cos(0.2))} \frac{-\mathrm{d}u}{u}\\ &=\Big[-\log|u|\Big]_{\log(\cos(0.1))}^{\log(\cos(0.2))} \\ &= -\log|\log(\cos 0.2)|+\log|\log(\cos 0.1)|\\ &={\log\left|\frac{\log(\cos(0.1))}{\log(\cos(0.2))}\right|}\\&= \log\left(\frac{\log(\cos(0.1))}{\log(\cos(0.2))}\right) \end{aligned}$$

Things to notice: the integrand is only defined when $\log(\cos x)$ exists AND is nonzero. So, for instance, it is not defined when $x=0$, because then $\log\cos x = \log1=0$, and we can't divide by zero.

In the final simplification, since $0.1$ and $0.2$ are between 0 and $\pi/2$, the cosine term is positive but less than one, so $\log(\cos 0.1)$ and $\log(\cos 0.2)$ are both negative; then their quotient is positive, so we can drop the absolute value signs.

Using the base change formula, we can also write the final answer as $\log\left(\log_{\cos (0.2)}\cos(0.1)\right)$.

(a) If we expand the integrand, one part of it is quite familiar–a portion of a circle. So, we split the specified integral in two.

$$\int_0^3(x+1)\sqrt{9-x^2} \ \mathrm{d}x = \int_0^3\sqrt{9-x^2} \ \mathrm{d}x +\int_0^3x\sqrt{9-x^2} \ \mathrm{d}x$$

The first piece represents the area above the $x$–axis and below the curve $y=\sqrt{9-x^2}$, i.e. $x^2+y^2=9$, with $0\le x\le 3$. That's the area of one quadrant of a disk of radius $3$. So

$$\int_0^3\sqrt{9-x^2} \ \mathrm{d}x =\frac{1}{4}(\pi \cdot 3^2) = \frac{9}{4}\pi$$

For the second part, we substitute $u=9-x^2$, $\mathrm{d}u= -2x\,\mathrm{d}x$. Note $u(0)=9$ and $u(3)=0$. So,

$$\begin{aligned} \int_0^3x\sqrt{9-x^2} \ \mathrm{d}x = \int_9^0\sqrt{u} \ \frac{\mathrm{d}u}{-2} =-\frac{1}{2}\left[\frac{u^{3/2}}{3/2}\right]_9^0 =-\frac{1}{2}\left[-\frac{27}{3/2}\right] =9 \end{aligned}$$

All together,

$$\int_0^3(x+1)\sqrt{9-x^2} \ \mathrm{d}x = \frac{9}{4}\pi + 9$$

(b) The integrand is of the form $N(x)/D(x)$ with $D(x)$ already factored and $N(x)$ of lower degree. We immediately look for a partial fractions decomposition:

$$\frac{4x+8}{(x-2)(x^2+4)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+4}.$$

Multiplying through by the denominator yields

$$\begin{aligned} 4x+8 &= A(x^2+4) + (Bx+C)(x-2) \tag{$*$} \end{aligned}$$

Setting $x=2$ we find:

$$\begin{aligned} 8 +8 = A(4 + 4) + 0 \implies 16 =8A \implies A = 2 \end{aligned}$$

Substituting $A=2$ in $(*)$ gives

$$\begin{aligned} &&4x+8 &= A(x^2+4) + (Bx+C)(x-2) \\ &\implies & -2x^2+4 x &=(x-2)(Bx + C) \\ &\implies\qquad & (-2x)(x-2)&=(Bx + C)(x-2) \\ & \implies& B& =-2,~C=0 \end{aligned}$$

So we have found that $A=2$, $B=-2$, and $C=0$. Therefore

$$\begin{aligned} \int\frac{4x+8}{(x-2)(x^2+4)}\ \mathrm{d}x &= \int \bigg( \frac{{red}2}{x-2} {red}- \frac{{red}2x}{x^2+4} \bigg) \,\mathrm{d}x \\ &= 2 \log |x-2| - \log (x^2 + 4) + C \end{aligned}$$

Here the second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution $u=x^2+4$, $\mathrm{d}u=2x\,\mathrm{d}x$.

(c) The given integral is improper, but only because of its infinite limits of integration. (The integrand is continuous for all real numbers.) So, we'll have to take two limits. Before we do that, though, let's find the antiderivative. We would like to use the substitution $u=e^x$, $\mathrm{d}u=e^x~\mathrm{d}x$. That is, $\frac{1}{u}~\mathrm{d}u = \mathrm{d}x$.

$$\begin{aligned} \int \frac{1}{e^x+e^{-x}}~\mathrm{d}x&=\int \frac{1}{u(u+\frac{1}{u})}~\mathrm{d}u = \int\frac{1}{u^2+1} = \arctan u+C \\&=\arctan(e^x)+C \\ \text{Now we can deal with the limits of integration.} \\ \int _{ -\infty}^\infty\frac{1}{e^x+e^{-x}}~\mathrm{d}x&= \int _{ -\infty}^0\frac{1}{e^x+e^{-x}}~\mathrm{d}x+ \int _{ 0}^\infty\frac{1}{e^x+e^{-x}}~\mathrm{d}x\\&= \lim_{a \to -\infty}\left[\int_{a}^0\frac{1}{e^x+e^{-x}}~\mathrm{d}x\right]+ \lim_{b \to \infty}\left[\int_{ 0}^b\frac{1}{e^x+e^{-x}}~\mathrm{d}x\right]\\&= \lim_{a \to -\infty}\big[\arctan(e^x)\big]_a^0+ \lim_{b \to \infty}\big[\arctan(e^x)\big]_0^b\\&= \lim_{a \to -\infty}\big[\arctan(e^0)-\arctan(e^a)\big]+ \lim_{b \to \infty}\big[\arctan(e^b)-\arctan(e^0)\big]\\&= \lim_{a \to -\infty}\big[-\arctan(e^a)\big]+ \lim_{b \to \infty}\big[\arctan(e^b)\big]\\ &=-\arctan(0)+\frac{\pi}{2} = \frac{\pi}{2} \end{aligned}$$