Question

int {2x²/(x-1)(x-2)(x-3)}dx​

Answer 1

EXPLANATION.

⇒ ∫2x².dx/(x - 1)(x - 2)(x - 3).

As we know that,

By applying the partial fraction method, we get.

Partial fraction is apply only  coefficient of Denominator > coefficient of numerator.

⇒ 2x²/(x - 1)(x - 2)(x - 3) = A/(x - 1) + B/(x - 2) + C/(x - 3).

Taking L.C.M in equation, we get.

⇒ 2x² = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2).

Put the value of x = 2 in equation, we get.

⇒ 2(2)² = A(2 - 2)(2 - 3) + B(2 - 1)(2 - 3) + C(2 - 1)(2 - 2).

⇒ 2(4) = 0 + B(1)(-1) + 0.

⇒ 8 = -B.

⇒ B = -8.

Put the value of x = 3 in equation, we get.

⇒ 2(3)² = A(3 - 2)(3 - 3) + B(3 - 1)(3 - 3) + C(3 - 1)(3 - 2).

⇒ 2(9) = 0 + 0 + C(2)(1).

⇒ 18 = 2C.

⇒ C = 9.

Put the value of x = 1 in equation, we get.

⇒ 2(1) = A(1 - 2)(1 - 3) + B(1 - 1)(1 - 3) + C(1 - 1)(1 - 2).

⇒ 2 = A(-1)(-2) + 0 + 0.

⇒ 2 = 2A.

⇒ A = 1.

Put the value in main equation, we get.

⇒ ∫2x²dx/(x - 1)(x - 2)(x - 3) = ∫A. dx/(x - 1) + ∫B. dx/(x - 2) + ∫C. dx/(x - 3).

⇒ ∫1.dx/(x - 1) + ∫-8.dx/(x - 2) + ∫9.dx/(x - 3).

⇒ ㏑(x - 1) - 8㏑(x - 2) + 9㏑(x - 3) + c.

                                                                                                                                     

MORE INFORMATION.

(1) = ∫p sin(x) + q cos(x)/a sin(x) + b cos(x) dx

(2) = ∫p sin(x)/a sin(x) + b cos(x).dx

(3) = ∫q cos(x)/a sin(x) + b cos(x).dx.

For their integration, we first express Nr. as follows.

Nr = A(Dr) + B(derivative of Dr).

Then integral = Ax + B ㏒(Dr) + c.

Answer 2

✬ TO SOLVE.

• ❍ ∫2x².dx/(x - 1)(x - 2)(x - 3)

✬ FORMULA USED.

• ❍ Partial fraction method = Partial fraction is apply only coefficient of Denominator > coefficient of numerator.

✬ SOLUTION.

➯ 2x²/(x - 1)(x - 2)(x - 3) = A/(x - 1) + B/(x - 2) + C/(x - 3)

☯ Then, take the L.C.M ,

➯ 2x² = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)

☯ Then, put the value of x = 2 ,

➯ 2(2)² = A(2 - 2)(2 - 3) + B(2 - 1)(2 - 3) + C(2 - 1)(2 - 2)

➯ 2(4) = 0 + B(1)(-1) + 0

➯ 8 = -B

➯ B = -8

☯ Then, put the value of x = 3,

➯ 2(3)² = A(3 - 2)(3 - 3) + B(3 - 1)(3 - 3) + C(3 - 1)(3 - 2)

➯ 2(9) = 0 + 0 + C(2)(1)

➯ 18 = 2C

➯ C = 9

☯ Then, put the value of x = 1,

➯ 2(1) = A(1 - 2)(1 - 3) + B(1 - 1)(1 - 3) + C(1 - 1)(1 - 2).

➯ 2 = A(-1)(-2) + 0 + 0.

➯ 2 = 2A.

➯ A = 1.

☯ Then,

➯ ∫2x²dx/(x - 1)(x - 2)(x - 3) = ∫A. dx/(x - 1) + ∫B. dx/(x - 2)+ ∫C. dx/(x - 3)

➯ ∫1.dx/(x - 1) + ∫-8.dx/(x - 2) + ∫9.dx/(x - 3)

➯ ㏑ (x - 1) - 8 ㏑ (x - 2) + 9 ㏑ (x - 3) + c                                                                                                                                   

✬ Hope it helps u ✬