부분분수 적분

$x \cosh (x)-\sinh (x)+C$

$\sinh^{-1} x +C=\ln\big(x+\sqrt{x^2+1}\big)+C$

$\cosh^{-1} x/3 +C=\ln\big(x+\sqrt{x^2-9}\big)+C$

$\left\{\begin{array}{ccc} \frac13\tanh^{-1}\left(\frac x3\right)+C & & x^2<9 \\ \\ \frac13\coth^{-1}\left(\frac x3\right)+C & & 9<x^2 \end{array}\right. = \frac12\ln |x+1| - \frac12\ln |x-1|+C$

$\cosh^{-1} (x^2/2) + C = \ln (x^2+\sqrt{x^4-4})+C$

$2/3\sinh^{-1} x^{3/2} + C = 2/3\ln (x^{3/2}+\sqrt{x^3+1})+C$

$\frac{1}{16}\tan^{-1}(x/2)+\frac{1}{32}\ln |x-2|+\frac1{32}\ln|x+2|+C$

$\ln x- \ln|x+1|+C$

$\tan^{-1}(e^x)+C$

$x\sinh^{-1}x-\sqrt{x^2+1}+C$

$x\tanh^{-1}x+1/2\ln|x^2-1|+C$

$\tan^{-1}(\sinh x)+C$

$0$

$3/2$

$2$

$\sinh^{-1} 2 =\ln\big(2+\sqrt{5}\big)\approx 1.444$

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution $u=(ax+b)$. It's less clear how to find the antiderivative of (2).

The integrand is a rational function, so it's a candidate for partial fraction. We quickly rule out any obvious substitution or integration by parts, so we go ahead with the decomposition.

We start by expressing the integrand, i.e. the fraction $\frac{1}{x+x^2}=\frac{1}{x(1+x)}$, as a linear combination of the simpler fractions $\frac{1}{x}$ and $\frac{1}{x+1}$ (which we already know how to integrate). We will have

$$\begin{aligned} \frac{1}{x+x^2}=\frac{1}{x(1+x)} = \frac{a}{x} + \frac{b}{x+1} =\frac{a(x+1) +bx}{x(1+x)} \end{aligned}$$

The fraction on the left hand side is the same as the fraction on the right hand side if and only if the numerator on the left hand side, which is $1 = 0x + 1$, is equal to the numerator on the right hand side, which is $a(x+1)+bx = (a+b)x +a$. This in turn is the case if and only of $a=1$ (i.e. the constant terms are the same in the two numerators) and $a+b=0$ (i.e. the coefficients of $x$ are the same in the two numerators). So $a=1$ and $b=-1$. Now we can easily evaluate the integral

$$\begin{aligned} \int_1^2 \frac{\mathrm{d}x}{x+x^2} &=\int_1^2 \frac{\mathrm{d}x}{x(x+1)} =\int_1^2 \Big(\frac{1}{x}-\frac{1}{x+1}\Big)\,\mathrm{d}x =\Big[\log x-\log(x+1)\Big]_1^2\\ &=\log 2-\log\frac{3}{2} =\log\frac{4}{3} \end{aligned}$$

The integrand is a rational function with no obvious substitution, so we'll use a partial fraction decomposition.

• Since the numerator has strictly smaller degree than the denominator, we don't need to start off with a long division.
• We do, however, need to factor the denominator. We can immediately pull out $x^2$; the remaining part is $x^2-2x+1 = (x-1)^2$.
• Now we can perform our partial fraction decomposition.

$$\begin{aligned} \frac{x^2+2x-1}{x^4-2x^3+x^2}&= \frac{x^2+2x-1}{x^2(x-1)^2} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2} \\ \text{Multiply both sides by the original denominator.} \\ x^2+2x-1&=Ax(x-1)^2+B(x-1)^2+Cx^2(x-1)+Dx^2\tag{1} \\ \text{To be sneaky, we set $x=0$, and find:} \\ -1&=B \\ \text{We also set $x=1$, and find:} \\ 2&=D \\ \text{We use $B$ and $D$ to simplify Equation (1).} \\ x^2+2x-1&=Ax(x-1)^2 -1(x-1)^2+Cx^2(x-1)+ 2x^2\\ 0&=Ax(x-1)^2+Cx^2(x-1)\\ &=x(x-1)[(A+C)x-A]\\ \text{So,}\qquad 0&=(A+C)x-A \end{aligned}$$

That is, $A=C=0$.

• Now we can evaluate our integral.

$$\begin{aligned} \int \frac{x^2+2x-1}{x^4-2x^3+x^2}~\mathrm{d}x&=\int \left(\frac{-1}{x^2}+\frac{2}{(x-1)^2}\right)~\mathrm{d}x\\ &=\frac{1}{x}-\frac{2}{x-1}+C \end{aligned}$$

Our integrand is a rational function with no obvious substitution, so we'll use the method of partial fractions.

• The degree of the numerator is less than the degree of the denominator.
• We need to factor the denominator. The first two terms have the same ratio as the last two terms.

$$\begin{aligned} 2x^3-x^2-8x+4&=x^2(2x-1)-4(2x-1) \\ &= (x^2-4)(2x-1)\\ &=(x-2)(x+2)(2x-1) \end{aligned}$$

• Now we find our partial fraction decomposition.

$$\begin{aligned} \frac{ 3x^2-4x-10}{2x^3-x^2-8x+4}&=\frac{ 3x^2-4x-10}{(x-2)(x+2)(2x-1)}= \frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{2x-1} \\ \text{Multiply both sides by the original denominator.} \\ 3x^2-4x-10&=A(x+2)(2x-1)+B(x-2)(2x-1)+C(x-2)(x+2) \\ \text{Distinct linear factors is the best possible scenario for the sneaky method. First, let's set $x=2$.} \\ 3(4)-4(2)-10&=A(4)(3)+B(0)+C(0)\\ A&=-\frac{1}{2} \\ \text{Now, let $x=-2$.} \\ 3(4)-4(-2)-10&=A(0)+B(-4)(-5)+C(0)\\ B&=\frac{1}{2} \\ \text{Finally, let $x=\frac{1}{2}$.} \\ \frac{3}{4}-2-10&=A(0)+B(0)+C\left(-\frac{3}{2}\right)\left(\frac{5}{2}\right)\\ C&=3 \end{aligned}$$

• Now we can evaluate our integral in its new form.

$$\begin{aligned} \int \frac{ 3x^2-4x-10}{2x^3-x^2-8x+4}~\mathrm{d}x&=\int \left(\frac{-1/2}{x-2}+\frac{1/2}{x+2}+\frac{3}{2x-1}\right) ~\mathrm{d}x\\ &=-\frac{1}{2}\log|x-2| + \frac{1}{2}\log|x+2| + \frac{3}{2}\log|2x-1|+C \\&=\frac{1}{2}\log\left| \frac{x+2}{x-2} \right| + \frac{3}{2}\log|2x-1|+C \end{aligned}$$