Integration By Partial Fractions

The numerator's degree exceeds the denominator's, so we divide first. Performing polynomial long division ofx^{3}$by$x^{2}-3x+2$ gives quotient $x+3$ and remainder $7x-6x^{2}-3x+2=(x-1)(x-2)$, since $1$and$2xA=-1$,$B=8x$, so we evaluate at the roots $x=1$and$x=2x=1x=2\displaystyle\int \frac{dx}{a^2 - x^2}a^{2}-x^{2}=(a-x)(a+x)x=a$and$x=-au^{2}-a^{2}=(u-a)(u+a)\displaystyle\int \frac{x + 4}{x^2(x^2 + 2x + 2)}\, dxx^{2}$ and the quadratic $x^{2}+2x+2$, whose discriminant $4-8=-4x=0xx=1x=-1x^{3}B=2x^{2}+2x+2=(x+1)^{2}+1$ and rewriting the numerator as $\tfrac{3}{2}x+1=\tfrac{3}{4}(2x+2)-\tfrac{1}{2}\displaystyle\int \frac{2x^3 + 3x^2 + 5x + 3}{(x^2 + 1)^2}\, dxxC$ is the constant of integration. For $\displaystyle\int \frac{3x}{(x^2+1)^2}\,dx$, we use the substitution $u=x^{2}+1$,so$du=2x\,dxDax^{2}+bx+c$ (which is $2ax+bn=1b^{2}-4ac<0$, the quantity $\dfrac{4ac-b^{2}}{4a^{2}}du=dxu=k\tan\theta$,so$du=k\sec^{2}\theta\,d\theta$and$u^{2}+k^{2}=k^{2}\sec^{2}\thetan=1n\geq2$, the integral of $\cos^{2n-2}\theta\displaystyle\int \frac{dx}{(x^2+1)^2}a=1$,$b=0$,$c=1$,$k=1$, and the square is already complete. We set $x=\tan\theta$,so$dx=\sec^{2}\theta\,d\theta$and$x^{2}+1=\sec^{2}\theta\cos^{2}\theta=\tfrac{1}{2}(1+\cos2\theta)x$. Since $x=\tan\theta$, we have $\theta=\arctan x$, and from the right triangle with opposite side $x$ and adjacent side $1$, hypotenuse $\sqrt{x^{2}+1}P(x)/Q(x)$ be a proper rational function in lowest terms. Suppose $(x-a)$ appears exactly $m$ times in $Q(x)$, so that $Q(x)=(x-a)^{m}S(x)$ where $S(a)\neq0A=P(a)/S(a)\neq0A$ be any constant and subtract $A/(x-a)^{m}$ from $P(x)/Q(x)(x-a)$ to cancel, $x=aS(a)\neq0$, this division is valid. Write $A_m=P(a)/S(a)$. Since the original fraction is in lowest terms, $P(a)\neq0$ (otherwise $(x-a)$ would be a common factor of $P$and$Q$), which ensures $A_m\neq0$. After canceling the factor of $(x-a)K_1(x)=[P(x)-A_m S(x)]/(x-a)K_1(x)/[(x-a)^{m-1}S(x)]$, which has exactly the same form with $m$ replaced by $m-1$. Setting $x=a$ shows $K_1(a)/S(a)$ is well-defined and nonzero; call it $A_{m-1}$. Subtracting $A_{m-1}/(x-a)^{m-1}$ and canceling $(x-a)K_2(x)=[K_1(x)-A_{m-1}S(x)]/(x-a)$ is again a polynomial and the fraction is again proper. Repeating the Lemma $m$ times in total, one step per power of $(x-a)$, extracts each constant $A_m,A_{m-1},\ldots,A_1$ in succession and exhausts the factor $(x-a)^{m}P(x)/Q(x)$ be a proper rational function in lowest terms. Suppose $x^{2}+px+q$ (where $p^{2}-4q<0$) appears exactly $m$ times in $Q(x)$, so that $Q(x)=(x^{2}+px+q)^{m}S(x)$ where $S(x)$ is not divisible by $x^{2}+px+qA$and$B(x^{2}+px+q)$ to cancel from the numerator, $P(x)-(Ax+B)S(x)$ must be divisible by $(x^{2}+px+q)$. To find $A$and$B$ that make this happen, we divide $P(x)$and$S(x)$ by the quadratic and record the remainders (each remainder has degree at most $1$, since the divisor has degree $2P(x)-(Ax+B)S(x)(x^{2}+px+q)x^{2}+px+qx^{2}+px+q$. Expanding $(Ax+B)(\lambda x+\mu)x^{2}+px+q$ and require the remainder to vanish. Performing the division: the leading term $-A\lambda x^{2}$ is absorbed by $-A\lambda$ times $x^{2}+px+q(-A\lambda)(x^{2}+px+q)=-A\lambda x^{2}-pA\lambda x-qA\lambda(x^{2}+px+q)A$and$B$. Since $L$and$M$ cannot both be zero, the equations are non-homogeneous. We claim that $A$and$B\lambda$and$\mu$ are not both zero: if they were, $S(x)$ would be divisible by $x^{2}+px+q\lambda\neq0$, factor out $\lambda^{2}x^{2}+px+q$ evaluated at $x=-\mu/\lambda$. Since $x^{2}+px+q$ is irreducible (its discriminant $p^{2}-4q\lambda=0$, then $\mu\neq0$, and the determinant reduces to $\mu^{2}>0A$and$BA$and$BA=0$and$B=0$. Equations (i) and (ii) then reduce to $L=0$and$M=0$.But$L$and$M$ are the coefficients of the remainder when $P(x)$ is divided by $x^{2}+px+q$. A zero remainder means $P(x)$ is exactly divisible by $x^{2}+px+q$. That makes $x^{2}+px+q$ a common factor of $P(x)$and$Q(x)$, contradicting the assumption that $P(x)/Q(x)$ is in lowest terms. So $A$and$B\dfrac{-3x^{2}-2x+3}{x^{3}+x^{2}}x^{3}+x^{2}=x^{2}(x+1)$. The factor $x$ is repeated twice. This means we need terms for both $x$and$x^{2}x^{2}(x+1)x=0x=-1x^{2}x^{2}$ on the left side is $-3$. On the right side, the $x^{2}$ terms come from $Ax^{2}$and$Cx^{2}C=2\dfrac{x+3}{(x-1)^{2}(x+2)}(x-1)x=1x=-2x=0B=4/3$and$C=1/9\dfrac{x^{2}+1}{x(x-1)^{2}}x=0x=1x^{2}1x^{2}$. The right side has $Ax^{2}+Bx^{2}A=11/(x-1)\dfrac{3x-1}{(x+2)^{3}}(x+2)x=-2x^{2}x^{2}$ term, so its coefficient is $0$. The right side has $Ax^{2}x3x$. The right side has $2Ax+Bx$, but we already know $A=0\displaystyle\int\frac{2x+1}{(x^{2}+2x+5)^{2}}\,dxx^{2}+2x+5=(x+1)^{2}+4$. We substitute $t=x+1$,so$dx=dtu=t^{2}+4$,so$du=2t\,dt\displaystyle I=\int\frac{dt}{(t^{2}+4)^{2}}$. We substitute $t=2\tan\theta$,so$dt=2\sec^{2}\theta\,d\theta(t^{2}+4)^{2}=16\sec^{4}\theta\cos^{2}\theta=\dfrac{1+\cos2\theta}{2}\sin2\theta=2\sin\theta\cos\thetat$. Since $t=2\tan\theta$, we have $\tan\theta=t/2$,so$\theta=\arctan(t/2)$. We read $\sin\theta$and$\cos\theta$ from a right triangle with opposite side $t$, adjacent side $2$, and hypotenuse $\sqrt{t^{2}+4}t=x+1\displaystyle\int\frac{dx}{x^{2}-4}a=2$, or decompose directly. Writing $\dfrac{1}{(x-2)(x+2)}=\dfrac{A}{x-2}+\dfrac{B}{x+2}$ gives $1=A(x+2)+B(x-2)$. Setting $x=2$ gives $A=\frac{1}{4}$, and setting $x=-2$ gives $B=-\frac{1}{4}\displaystyle\int\frac{dx}{x^{2}-5x+6}x^{2}-5x+6=(x-2)(x-3)$. Then $1=A(x-3)+B(x-2)$. Setting $x=3$ gives $B=1$; setting $x=2$ gives $A=-1\displaystyle\int\frac{5x-2}{x^{2}-4}\,dx\dfrac{5x-2}{(x-2)(x+2)}=\dfrac{A}{x-2}+\dfrac{B}{x+2}$,so$5x-2=A(x+2)+B(x-2)x=2$:$8=4A$,so$A=2$. Setting $x=-2$:$-12=-4B$,so$B=3\displaystyle\int\frac{x+1}{x^{2}-x-6}\,dxx^{2}-x-6=(x-3)(x+2)$. Then $x+1=A(x+2)+B(x-3)x=3$:$4=5A$,so$A=\frac{4}{5}$. Setting $x=-2$:$-1=-5B$,so$B=\frac{1}{5}\displaystyle\int\frac{dx}{x(x+1)}1=A(x+1)+Bx$. Setting $x=0$ gives $A=1$; setting $x=-1$ gives $B=-1\displaystyle\int\frac{dx}{x^{2}(x+1)}xx=0$ gives $B=1$. Setting $x=-1$ gives $C=1$. Comparing coefficients of $x^{2}$:$0=A+C$,so$A=-1\displaystyle\int\frac{dx}{x^{3}-x}x^{3}-x=x(x-1)(x+1)x=0$:$-A=1$,so$A=-1$. Setting $x=1$:$2B=1$,so$B=\frac{1}{2}$. Setting $x=-1$:$2C=1$,so$C=\frac{1}{2}\dfrac{1}{2}\ln\left|\dfrac{x^{2}-1}{x^{2}}\right|+C\displaystyle\int\frac{x^{2}+2}{(x+1)^{3}}\,dxu=x+1$,so$x=u-1$and$x^{2}+2=(u-1)^{2}+2=u^{2}-2u+3u=x+a\displaystyle\int\frac{x^{2}+x+1}{x(x^{2}+1)}\,dxx^{2}+1x=0$ gives $A=1$. Comparing coefficients of $x^{2}$:$1=A+B$,so$B=0$. Comparing coefficients of $x$:$1=C\displaystyle\int\frac{2x^{2}-x+4}{x^{3}+4x}\,dxx^{3}+4x=x(x^{2}+4)$,and$x^{2}+4x=0$:$4A=4$,so$A=1$. Comparing $x^{2}$ coefficients: $2=A+B$,so$B=1$. Comparing $x$ coefficients: $-1=C\displaystyle\int\frac{dx}{(x+1)(x^{2}+1)}x=-1$:$2A=1$,so$A=\frac{1}{2}$. Comparing $x^{2}$ coefficients: $0=A+B$,so$B=-\frac{1}{2}$. Comparing constants: $1=A+C$,so$C=\frac{1}{2}\displaystyle\int\frac{x^{4}}{x^{2}-1}\,dx(x^{2}+1)(x^{2}-1)=x^{4}-1$, we have $x^{4}=(x^{2}+1)(x^{2}-1)+1a=1\displaystyle\int\frac{x^{3}+x+2}{x^{2}+2}\,dxx^{3}+x+2=x(x^{2}+2)+(-x+2)\displaystyle\int\frac{dx}{x(x^{2}+1)^{2}}x^{2}+1x=0$ gives $A=1x^{4}$:$A+B=0$,so$B=-1$.$x^{3}$:$C=0$.$x^{2}$:$2A+B+D=0$,so$D=-1$.$x$:$C+E=0$,so$E=0u=x^{2}+1\displaystyle\int\frac{dx}{x^{4}-1}x^{4}-1=(x-1)(x+1)(x^{2}+1)$, where $x^{2}+1x=1$:$4A=1$,so$A=\frac{1}{4}$. Setting $x=-1$:$-4B=1$,so$B=-\frac{1}{4}$. Comparing $x^{3}$ coefficients: $A+B+C=0$,so$C=0$. Comparing constants: $A-B-D=1$,so$\frac{1}{4}+\frac{1}{4}-D=1$and$D=-\frac{1}{2}\displaystyle\int\frac{e^{x}}{e^{2x}-1}\,dxu=e^{x}$,so$du=e^{x}\,dx$and$e^{2x}=u^{2}a=1\displaystyle\int\frac{\cos x}{\sin^{2}x+\sin x-2}\,dxu=\sin x$,so$du=\cos x\,dx1=A(u-1)+B(u+2)$. Setting $u=1$:$3B=1$,so$B=\frac{1}{3}$. Setting $u=-2$:$-3A=1$,so$A=-\frac{1}{3}\displaystyle\int\frac{dP}{P(K-P)}$, where $K>0$ is the carrying capacity and $0<P<K\dfrac{1}{P(K-P)}=\dfrac{A}{P}+\dfrac{B}{K-P}$,so$1=A(K-P)+BPP=0$:$AK=1$,so$A=\frac{1}{K}$. Setting $P=K$:$BK=1$,so$B=\frac{1}{K}0 a n d K-P$ positive. Solving this relation for $Pxx=-1$ in the cubic gives $-1+5-17+13=0$,so$(x+1)x^{2}+4x+13$ has discriminant $16-52=-36<0x=0$:$26=13A$,so$A=2$. Setting $x=-1$:$-3+11-44+26=-10$ on the left, and on the right $B(-1)(1-4+13)=-10B$,so$B=1$. Comparing coefficients of $x^{3}$:$3=A+B+C=2+1+C$,so$C=0$. Comparing constants we have already used $x=0$; instead compare coefficients of $x^{2}$: on the right, $A(4+1)+B(4)+D=5A+4B+D$, which must equal $11$, giving $10+4+D=11$and$D=-3x^{2}+4x+13=(x+2)^{2}+9$, gives the integral quoted earlier: