Exercises — Partial fractions — linear, repeated, irreducible quadratic factors

Do distinct linear factors hain. Har linear factor $(x-a)$ apne upar ek single constant contribute karta hai: $\frac{5x-2}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}.$ Count check: $\mathrm{deg}Q=2$, aur hamare paas $2$ unknowns $A,B$ hain ✓.

$(x+1{)}^3$ order $3$ ka repeated linear factor hai: hume har power $3$ se $1$ tak neeche run karni padegi (akela ek term us numerator ko represent nahi kar sakta jo secretly $(x+1)$ mein quadratic ho sakta hai). Plus akele $(x-5)$ ke liye ek block: $\frac{7}{(x+1{)}^3(x-5)}=\frac{A_1}{x+1}+\frac{A_2}{(x+1{)}^2}+\frac{A_3}{(x+1{)}^3}+\frac{B}{x-5}.$ Count check: $\mathrm{deg}Q=4$, unknowns $A_1,A_2,A_3,B=4$ ✓.

Irreducible? $x^2+0x+9$ ka discriminant $0^2-4(1)(9)=-36<0$ hai, toh koi real roots nahi — haan, irreducible. Ek irreducible quadratic ko linear top $Bx+C$ milta hai (numerator degree denominator se ek kam). Proper? $\mathrm{deg}P=3$, $\mathrm{deg}Q=1+2=3$. Kyunki $\mathrm{deg}P=\mathrm{deg}Q$ hai, yeh improper hai — akela template kaafi nahi; pehle divide karna hoga. Lekin remainder ka template yeh hai: $\frac{x^3+1}{(x-2)(x^2+9)}=(\text{division se constant})+\frac{A}{x-2}+\frac{Bx+C}{x^2+9}.$ Count check (proper part): $\mathrm{deg}Q=3$, unknowns $A,B,C=3$ ✓.

Template: $\frac{5x-2}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}.$ $A$ ke liye cover-up: $(x-3)$ cover karo, $x=3$ set karo: $A=\frac{5(3)-2}{3+4}=\frac{13}{7}.$ (the $B$ term carried $(x-3)\to 0$.) $B$ ke liye cover-up: $(x+4)$ cover karo, $x=-4$ set karo: $B=\frac{5(-4)-2}{-4-3}=\frac{-22}{-7}=\frac{22}{7}.$ Integrate (har $\frac{A}{x-a}\to A\ln |x-a|$): $\int =\frac{13}{7}\ln |x-3|+\frac{22}{7}\ln |x+4|+C.$

$(x-2{)}^2$ repeated ⇒ template $\frac{3x-1}{(x-2{)}^2}=\frac{A}{x-2}+\frac{B}{(x-2{)}^2}.$ $B$ (top power par cover-up): $x=2$ set karo: $B=3(2)-1=5.$ $A$ (matching): denominators clear karo: $3x-1=A(x-2)+B$. $x$ ka coefficient: $A=3$. (Check constant: $-2A+B=-6+5=-1$ ✓.) Integrate: doosra piece power rule use karta hai, log nahi (exponent $-2\ne -1$): $\int =3\ln |x-2|-\frac{5}{x-2}+C.$

Cover up karne ke liye koi linear factor nahi hai — parent rule se split karo "log the derivative, arctan the square." $x^2+4$ ka derivative $2x$ hai, toh numerator engineer karo: $4x+1=2\cdot (2x)+1.$ $\int \frac{4x+1}{x^2+4}dx=2\int \frac{2x}{x^2+4}dx+\int \frac{1}{x^2+4}dx.$ Pehla integral substitution $u=x^2+4$ se: $2\ln (x^2+4)$. Doosra $d^2=4$, $d=2$ ke saath standard arctan hai: $\int \frac{1}{x^2+2^2}dx=\frac{1}{2}\arctan \frac{x}{2}$. $\int =2\ln (x^2+4)+\frac{1}{2}\arctan \frac{x}{2}+C.$

Template: $\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+2}.$ $B$ cover-up (repeat ki top power): $x=1$: $B=\frac{1+4}{1+2}=\frac{5}{3}.$ $C$ cover-up: $x=-2$: $C=\frac{-2+4}{(-3{)}^2}=\frac{2}{9}.$ $A$ — the analysis: $A$ ka koi clean cover-up nahi hai. Multiply out karne par, numerator ka $x^2$ coefficient wahi jagah hai jahan $A$ aur $C$ bina $B$ ke milte hain. Kyunki left numerator $x+4$ mein koi $x^2$ term nahi hai: $A+C=0\Rightarrow A=-C=-\frac{2}{9}.$ $x^2$ coefficient kyun choose karein? yeh $A$ ek line mein de deta hai $3\times 3$ system solve karne ki jagah. Integrate: $\int =-\frac{2}{9}\ln |x-1|-\frac{5}{3}\cdot \frac{1}{x-1}+\frac{2}{9}\ln |x+2|+C.$

Check irreducible: discriminant $2^2-4(5)=4-20=-16<0$ ✓. Denominator ka derivative $2x+2$ hai. Numerator engineer karo taaki derivative exactly appear ho: $x+3=\frac{1}{2}(2x+2)+2.$ $\int =\frac{1}{2}\int \frac{2x+2}{x^2+2x+5}dx+\int \frac{2}{x^2+2x+5}dx.$ Pehla piece: $\frac{1}{2}\ln (x^2+2x+5)$. Doosra: complete the square $x^2+2x+5=(x+1{)}^2+4$, toh $d=2$: $\int \frac{2}{(x+1{)}^2+2^2}dx=2\cdot \frac{1}{2}\arctan \frac{x+1}{2}=\arctan \frac{x+1}{2}.$ $\int =\frac{1}{2}\ln (x^2+2x+5)+\arctan \frac{x+1}{2}+C.$

Template: $\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$ (quadratic irreducible: discriminant $-4<0$). $A$ cover-up: $x=-1$: $A=\frac{2(1)-1+1}{(-1{)}^2+1}=\frac{2}{2}=1.$ $B,C$ by clearing: $2x^2+x+1=A(x^2+1)+(Bx+C)(x+1)$. $A=1$ ke saath: $2x^2+x+1=(1+B)x^2+(B+C)x+(1+C).$ Match $x^2:1+B=2\Rightarrow B=1$. Match const: $1+C=1\Rightarrow C=0$. Check $x$: $B+C=1$ ✓. Toh $\frac{2x^2+x+1}{(x+1)(x^2+1)}=\frac{1}{x+1}+\frac{x}{x^2+1}.$ Integrate (linear ke liye log, quadratic ke liye $u=x^2+1$ se half-log; $C=0$ hone se koi arctan nahi): $\int =\ln |x+1|+\frac{1}{2}\ln (x^2+1)+C.$

Proper? $\mathrm{deg}P=2=\mathrm{deg}Q$ ⇒ improper. Pehle long-divide karo: $\frac{x^2}{x^2-1}=1+\frac{1}{x^2-1}.$ Ab $x^2-1=(x-1)(x+1)$, dono linear: $\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.$ $x=1$ par $A$ cover-up: $A=\frac{1}{1+1}=\frac{1}{2}$. $x=-1$ par $B$ cover-up: $B=\frac{1}{-1-1}=-\frac{1}{2}$. Integrate (constant $1$ integrate hokar $x$ banega): $\int =x+\frac{1}{2}\ln |x-1|-\frac{1}{2}\ln |x+1|+C=x+\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C.$

Irreducible? discriminant $2^2-4(2)=4-8=-4<0$ ✓. $x^2+2x+2$ ka derivative $2x+2$ hai. $2x+5=(2x+2)+3$ likho. $\int =\int \frac{2x+2}{x^2+2x+2}dx+\int \frac{3}{x^2+2x+2}dx.$ Pehla = $\ln (x^2+2x+2)$. Doosra: complete the square $x^2+2x+2=(x+1{)}^2+1$, $d=1$: $3\int \frac{1}{(x+1{)}^2+1}dx=3\arctan (x+1).$ $\int =\ln (x^2+2x+2)+3\arctan (x+1)+C.$

Proper? $\mathrm{deg}P=3$, $\mathrm{deg}Q=2+2=4$, toh $3<4$ ✓ proper. Template (repeated linear dono powers run karta hai; irreducible quadratic ko $Bx+C$ milta hai): $\frac{x^3-2x+3}{(x-1{)}^2(x^2+1)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{Cx+D}{x^2+1}.$ Count: $\mathrm{deg}Q=4$, unknowns $A,B,C,D=4$ ✓. $B$ find karo (top power cover-up): $x=1$: $B=\frac{1-2+3}{1^2+1}=\frac{2}{2}=1.$ Denominators clear karo: $x^3-2x+3=A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1{)}^2.$ $B=1$ ke saath expand karo aur powers match karo:

• $x^3$: $A+C=1$
• $x^2$: $-A+1-2C+D=0$
• $x^1$: $A-2D+C=-2$ … (using $A-2C+...$ careful expansion below) Cleanly karo. $A(x-1)(x^2+1)=A(x^3-x^2+x-1)$. $(Cx+D)(x-1{)}^2=(Cx+D)(x^2-2x+1)=C{x}^3+(D-2C)x^2+(C-2D)x+D$. Sum coefficients (add $B(x^2+1)=x^2+1$):
• $x^3:A+C=1$
• $x^2:-A+(D-2C)+1=0$
• $x^1:A+(C-2D)=-2$
• $x^0:-A+D+1=3$ $x^0$ se: $D=A+2$. $x^3$ aur $x^1$ mein sub karo: $C=1-A$, aur $A+(1-A)-2(A+2)=-2\Rightarrow 1-2A-4=-2\Rightarrow -2A=1\Rightarrow A=-\frac{1}{2}.$ Phir $C=1-(-\frac{1}{2})=\frac{3}{2}$, $D=-\frac{1}{2}+2=\frac{3}{2}$. (Check $x^2$: $-(-\frac{1}{2})+(\frac{3}{2}-3)+1=\frac{1}{2}-\frac{3}{2}+1=0$ ✓.) $\frac{x^3-2x+3}{(x-1{)}^2(x^2+1)}=\frac{-\frac{1}{2}}{x-1}+\frac{1}{(x-1{)}^2}+\frac{\frac{3}{2}x+\frac{3}{2}}{x^2+1}.$