Math Twelve

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`

12-math-home


Reference: