Integrals NCERT Solutions
NCERT Exercise: 7.5
Question: 1. `x/((x+1)(x+2)`
Solution: `int x/((x+1)(x+2)``dx`
The integral `x/((x+1)(x+2)` is a proper rational function.
∴ `x/((x+1)(x+2)` `=` `A/(x+1)` `+` `B/(x+2`
`rArr` `x = A (x + 2) + B (x + 1)`
`rArr x = (A+B) x + (2A + B)`
Equating coefficients of like terms on both sides, we have
`A + B = 1`
`2A + B = 0`
Subtracting (ii) from (iii), we have
`A = - 1`
Putting `A = -1` in equation (ii), we have
`B = 2`
Putting values of A and B in (i), we have
`x/((x+1)(x+2)` `= -1/(x+1) + 2/(x+2)`
∴ `int x/((x+1)(x+2)` `dx`
`= int [(-1)/(x+1) + 2/(x+2)]dx`
`=- int 1/(x+1) dx + 2 int 1/(x+2) dx`
`= - log |x+1| + 2 log |x+2| + C`
`= - log |x+1| + log (x+2)^2 + C`
`= log (x+2)^2 / (|x+1|) + C`
Question: 2. `1/(x^2 - 9)`
Solution:
`int 1/(x^2 - 9) dx` `=int 1/((x+3)(x - 3)` `dx`
The integrand `1/((x+3)(x-3)` is a proper rational function.
∴ `1/((x+3)(x-3)` `=A/(x+3)+ B/(x-3)` --- (i)
`rArr 1=A(x-3)+B(x+3)`
`rArr 1 = (A+B)x+(-3A+3B)`
Equating coefficient of like terms on both sides, we get
`A+B=0` ---- (ii)
`-3A+3B = 1` ----(iii)
Multiplying (ii) by 3 and adding with (ii), we get
`B=1/6`
Now, by putting the value of B in (ii), we get
`A=1/6`
Now after putting values of A and B in (i), we get
`1/((x+3)(x-3)``=-1/(6(x+3))+1/(6(x-3)`
∴`int 1/((x+3)(x - 3)``dx`
`=int[(-1)/(6(x+3))+1/(6(x-3))]dx`
`=-1/6int1/(x+3)dx+1/6int1/(x-3)dx`
`=-1/6log|x+3|+1/6log|x-3|+C`
`=1/6log|(x-3)/(x+3)|+CAnswer`
Question: 3. `int(3x-1)/((x-1)(x-2)(x-3))`
Solution: `int(3x-1)/((x-1)(x-2)(x-3))dx`
The integrand `(3x-1)/((x-1)(x-2)(x-3))` is a proper rational function.
∴ `(3x-1)/((x-1)(x-2)(x-3))``=A/(x-1)+B/(x-2)+C/(x-3)` ---(i)
`rArr 3x-1` `= A(x-2)(x-3)+B(x-1)(x-3)``+C(x-1)(x-2)`
`rArr 3x-1` `=A(x^2-5x+6)``+B(x^2-4x+3)``+C(x^2-3x+2)`
`rArr 3x-1``=(A+B+C)x^2``+(5A-4B-3C)x``+(6A+3B+2C)`
Equating coefficients of like terms on bothe sides, we get
`A+B+C=0` ---(ii)
`-5A-4B-3C=3` ----(iii)
`6A+3B+2C=-1` ---(iv)
By multiplying (ii) by 3 and adding with (iii), we get
`-2A-B=3` ---(v)
By multiplying (ii) by 2 and then subtracting from (iv), we get
`4A+B=-1` -----(vi)
Now, by adding (v) and (vi), we get
`A=1`
Now, by putting the values of A in (vi), we get
`B=-5`
Now, by putting the values of A and B in (ii), we get
`C=4`
Now, by putting the values of A,B and C in (i), we get
`(3x-1)/(((x-1)(x-2)(x-3)``=1/(x-1)-5/(x-2)+4/(x-3)`
∴ `int(3x-1)/((x-a)(x-2)(x-3))dx`
`=int [1/(x-1)-5/(x-2)+4(x-3)]dx`
`=int1/(x-1)dx-5int1/(x-2)dx+4int1/(x-3)dx`
`=log |x-1|-5log|x-2|+4log|x-3|+C`
Question: 4. `intx/((x-1)(x-2)(x-3))dx`
Solution:
`intx/((x-1)(x-2)(x-3))dx`
The integrand `x/((x-1)(x-2)(x-3))dx`is a proper rational function.
∴ `x/((x-1)(x-2)(x-3))dx`
`=A/(x-1)+B/(x-2)+C/(x-3)`
`rArrx=A(x-2)(x-3)+B(x-1)(x-3)` `+C(x-1)(x-2)`
`=A(x^2-5x+6)` `+B(x^2-4x+3)` `+C(x^2-3x+2)`
`=(A+B+C)x^2` `+(-5A-4B-3C)x` `+(6A+3B+2C)`
Equating coefficients of like terms on both sides, we have
`A+B+C=0`----(ii)
`-5A-4B-3C=1` ----(iii)
`6A+3B+2C=0` ----(iv)
Multiplying (ii) by 3 and adding with (iii), we have
`2A+B=-1` ----(v)
Multiplying (ii) by 2 and then subtracting from (iv), we have
`4A+B=0`
Subtracting (vi) from (v)
`A=1/2`
Putting values of A in (vi)
`B=-2`
Putting values of A and B in (ii)
`C=3/2`
Putting values of A,B and C in (i)
`x/((x-1)(x-2)(x-3))` `=1/(2(x-1))-2/(x-2)+3/(2(x-3))`
∴ `intx/((x-1)(x-2)(x-3))dx`
`=int[1/(2(x-1))-2/(x-2)+3/(2(x-3))]dx`
`=1/2int1/(x-1)dx-2int1/(x-2)dx` `+3/2int1/(x-3)dx`
`=1/2log|x-1|-2log|x-2|` `+3/2log|x-3|+C`
Question: 5. `2x/(x^2+3x+2)`
Solution:
`int2x/(x^2+3x+2)dx`
`=int2x/((x+1)(x+2))dx`
The integrand `2x/(x^2+3x+2)` is a proper function
`:. 2x/((x+1)(x+2))=A/(x+1)+B/(x+2)`
`=>x=A(x+2)+B(x+1)`
`=>x=(A+B)x+(2A+B)`
Equating coefficients of like terms on both sides, we get
`A+B=2` ---(ii)
`2A+B=0` ---(iii)
Subtracting (ii) from (iii)
`A=-2`
Putting the value of A in (ii)
`B=4`
Putting the values of A and B in (i)
`(2x)/((x+1)(x+2))=(-2)/(x+1)+4/(x+2)`
`:.int2x/((x+1)(x+2)dx` `=int[(-2)/(x+1)+4/(x+2)]dx`
`=-2int1/(x+1)dx+4int1/(x+2)dx`
`=-2log|x+1|+4log|x+2|+C`
Question: 6. `(1-x^2)/(x(1-2x))`
Solution:
`int(1-x^2)/(x(1-2x))dx`
`=int[1/2+(-1/2x+1)/(x(1-2x))]dx`
`=1/2intdx-1/2int(x-2)/(x(1-2x))dx`
`=x/2-1/2I_1` --(i)
Now, `I_1 = int(x-2)/(x(1-2x))dx`
The integrand`(x-2)/(x(1-2x))` is a proper rational function
`:.(x-2)/(x(1-2x))=A/x+B/(1-2x)` ---(ii)
`=>x-2=A(1-2x)+Bx`
`=>x-2(-2A+B)x+A`
Equating coefficients of like terms on both sides, we have
`-2A+B=1` ----(iii)
`A=-2` ----(iv)
Putting value of A in (iii), we get
`B=-3`
Putting the value of A and B in (ii). we get
`(x-2)/(x(1-2x))=(-2)/x-3/(1-2x)`
`:. int(x-2)/(x(1-2x))dx=int[(-2)/x-3/(1-2x)]dx`
`=-2int1/xdx-3int1/(1-2x)dx`
`=-2log|x|-3(log|1-2x|)/-2+C_1`
`=-2log|x|-3/2log|1-2x|+C_1`
putting the value of` 1_1(i)`
`int(1-x^2)/(x(1-2x))dx=x/2-1/2` `[-2log|x|-3/2log|1-2x|+C_1]`
`=x/2+log|x|-3/4log|1-2x|-(C_1)/2`
`=x/2+log|x|-3/4log|1-2x|-C`
Where `C=(C_1)/2`
Question:7.`x/((x^2+1)(x-1))`
Solution:
`intx/((x^2+1)(x-1))dx`
The integrand `x/((x^2+1)(x-1)` is a proper rational function.
∴ `x/((x^2+1)(x-1))=A/(x-1)+(Bx+C)/(x^2+1)`
`x=A(x^2+1)+(Bx+C)(x-1)`
`=>x=(A+B)x^2+(-B+C)x+(A-C)`
Equation coefficients of terms on both sides, we have
`A+B=0`
`-B+C=1`
`A-C=0`
Adding (ii) and (iii),
`A+C=1`
Adding (iv) and (v),
`A=1/2`
Putting value of A in (v),
PuttingValue of B in (ii),
`B=-1/2`
Putting values of A, B and C in (i),
`x/((x^2+1)(x-1))=1/(2(x-1))+` `(-x+1)/(2(x^2+1))`
∴`intx/((x^2+1)(x-1))dx` `=int[1/(2(x-1))-1/2 (x+1)/(x^2+1)]dx`
`=1/2int 1/(x-1)dx-1/2intx/(x^2+1)dx` `+1/2int1/(x^2+1)dx`
`=1/2int1/(x-1)dx-1/4int(2x)/(x^2+1)dx` `+1/2int1/(x^2+1)dx`
`=1/2log|x-1|-1/4log|x^2+1|+` `1/2tan^-1x+C`
Reference: