Evaluate the integral ∫ d x x ( x + 1 ) ( x + 2 )

Let I = ∫ d x x ( x + 1 ) ( x + 2 ) Consider 1 x ( x + 1 ) ( x + 2 ) = A x + B x + 1 + C x + 2 1 = A ( x + 1 ) ( x + 2 ) + B ( x ) ( x + 2 ) + c ( x ) ( x + 1 ) Put x = - 2 ⇒ 2 C = 1 ⇒ C = 1 / 2 . x = 0 ⇒ 2 A = 1 ⇒ A = 1 / 2 x = - 1 ⇒ - B = 1 Therefore, ∫ d x 2 ( x + 1 ) ( 2 + 2 ) = ∫ 1 2 x d x - ∫ d x x + 1 + ∫ 1 2 ( x + 2 ) d x I = 1 2 log x - log ( x + 1 ) + 1 2 log ( x + 2 )

Evaluate ∫ 3 x - 2 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x

Let I = ∫ 3 x - 2 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x Consider 3 x - 2 ( x - 1 ) ( x + 2 ) ( x - 3 ) = A ( x - 1 ) + B ( x + 2 ) + C ( x - 3 ) ⇒ 3 x - 2 = A ( x + 2 ) ( x - 3 ) + B ( x - 1 ) ( x - 3 ) + c ( x - 1 ) ( x + 2 ) x = 3 ⇒ 10 C = 7 ⇒ C = 7 / 10 x = - 2 ⇒ 15 B = - 3 ⇒ B = - 1 / 5 x = 1 ⇒ - 6 A = 1 ⇒ A = - 1 / 6 ∫ 3 x - 2 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x = - ∫ d x 6 ( x - 1 ) - ∫ d x 5 ( x + 2 ) + ∫ 7 d x 10 ( x - 3 ) = - 1 6 log ( x - 1 ) - 1 5 log ( x + 2 ) + 7 10 log ( x - 3 )

Evaluate the integral ∫ 7 x - 4 ( x - 1 ) 2 ( x + 2 ) d x

Let I = ∫ 7 x - 4 ( x - 1 ) 2 ( x + 2 ) d x Consider 7 x - 4 ( x - 1 ) 2 ( x + 2 ) = A ( x - 1 ) + B ( x - 1 ) 2 + C ( x + 2 ) 7 x - 4 = A ( x - 1 ) ( x + 2 ) + B ( x + 2 ) + ( x - 1 ) 2 7 x - 4 = A x 2 + A x - 2 A + B x + 2 B + C x 2 - 26 x + C Equating coefficients A + C = 0 A + B - 2 C = 7 - 2 A + 2 B + C = - 4 A = 2 C = - 2 and B = 1 ∫ 7 x - 4 ( x - 1 ) 2 ( x + 2 ) = ∫ 2 x - 1 d x + ∫ d x ( x - 1 ) 2 - 2 ∫ d x x + 2 I = 2 log | x + 1 | + - 1 x - 1 - 2 log ( x + 2 ) + C I = 2 log | x - 1 | | x + 2 | - 1 ( x - 1 ) + C

Evaluate the integral ∫ 1 ( x - a ) ( x - b ) ( x - c ) d x

Let I = ∫ 1 ( x - a ) ( x - b ) ( x - c ) d x Consider 1 ( x - a ) ( x - b ) ( x - c ) = A ( x - a ) + B ( x - b ) + C ( x - c ) 1 = A ( x - b ) ( x - c ) + B ( x - a ) ( x - c ) + C ( x - a ) ( x - b ) x = a ⇒ 1 = A ( a + b ) ( α - c ) ⇒ A = 1 ( a - b ) ( a - c ) x + b ⇒ 1 = ( b - a ) ( b - c ) ⇒ B = 1 ( b - a ) ( b - c ) x = c ⇒ 1 = C ( c - 1 ) ( c - b ) ⇒ C = 1 ( c - a ) ( c - b ) ∴ ∫ d x ( x - 4 ) ( x - b ) ( x - 1 ) = 1 ( x - b ) ( x - c ) ∫ d x x - a + 1 ( b - a ) ( b - c ) ∫ d x ( x - b ) + + 1 ( c - a ) ( c - b ) ∫ d x x - c = log ( x - a ) ( a - b ) ( a - c ) + log ( x - b ) ( b - a ) ( b - c ) + log ( x - c ) ( c - a ) ( c - b )

Evaluate the integral ∫ 2 x + 3 ( x + 3 ) x 2 + 4 d x

Let I = ∫ 2 x + 3 ( x + 3 ) x 2 + 4 d x Consider 2 x + 3 ( x + 3 ) x 2 + 4 = A x + 3 + B x + C x 2 + 4 2 x + 3 = A x 2 + 4 + ( B x + C ) ( x + 3 ) 2 x + 3 = A x 2 + 4 A + B x 2 + 3 B x + C x + 3 C 2 x + 3 = ( A + B ) x 2 + ( 3 B + C ) x + 4 A + 3 C Comparing coefficient of x 2 , x and constants, we get A + B = 0 ⇒ A = - B 3 B + C = 2 and 4 A + 3 C = 3 ⇒ - 3 A + C = 2 and 4 A + 3 C = 3 - 12 A + 4 C = 8 + 12 A + 9 C = 9 13 C = 17 Therefore, C = 17 13 , B = 3 13 , A = - 3 13 . ∴ I = - 3 13 ∫ d x x + 3 + ∫ 3 x + 17 13 x 2 + 4 d x = - 3 13 log ( x + 3 ) + 3 13 ∫ 2 x 2 x 2 + 4 d x + 17 13 ∫ d x x 2 + 4 Hence, I = - 3 13 log ( x + 3 ) + 3 26 log x 2 + 4 + 17 26 tan - 1 x 2 + C .

Evaluate the integral ∫ 2 x 2 + x + 1 ( x + 3 ) ( x - 2 ) 2 d x

Let I = ∫ 2 x 2 + x + 1 ( x + 3 ) ( x - 2 ) 2 d x Consider 2 x 2 + x + 1 ( x + 3 ) ( x - 2 ) 2 = A x + 3 + B x - 2 + C ( x - 2 ) 2 2 x 2 + x + 1 = A ( x - 2 ) 2 + B ( x + 3 ) ( x - 2 ) + c ( x + 3 ) ⇒ 2 x 2 + x + 1 = A x 2 + 4 - 4 x + B x 2 + x - 6 + C x + 3 C ⇒ 2 x 2 + x + 1 = x 2 ( A + B ) + x [ - 4 A + B + C ] + 4 A - 6 B + 3 C Equating the coefficients we get A + B = 2 , - 4 A + B + C = 1 , 4 A - 6 B + 3 C = 1 On solving we get A = 16 25 , B = 34 25 , C = 11 5 Therefore, I = 16 25 ∫ 1 x + 3 d x + 34 25 ∫ 1 x - 2 d x + 11 5 ∫ 1 ( x - 2 ) 2 d x I = 16 25 ∫ 1 x + 3 d x + 34 25 ∫ 1 x - 2 d x + 11 5 ∫ x - 2 - 2 d x I = 16 25 ln x + 3 + 34 25 ln x - 2 - 11 5 1 ( x - 2 ) + C

Evaluate the integral ∫ d x x 3 + 1 .

Let I = ∫ d x x 3 + 1 Consider 1 x 3 + 1 = A x + B x 2 - x + 1 + C x + 1 Multiplying both sides by x 2 - x + 1 ( x + 1 ) 1 = ( A x + B ) ( x + 1 ) + C x 2 - x + 1 x = - 1 ⇒ 1 = C ( - 1 ) 2 - - 1 + 1 C = 1 3 x = 0 ⇒ 1 = B + 1 3 ⇒ B = 2 3 When x = 1 ⇒ 1 = A + 2 3 ( 2 ) + 1 3 ⇒ A = - 1 3 1 x 3 + 1 = 1 3 1 x + 1 - 1 3 x - 2 x 2 - x + 1 We want 2 x - 1 in the numerator of the second term, therefore we much create a third term for the remaining -3: 1 x 3 + 1 = 1 3 1 x + 1 - 1 6 2 x - 1 x 2 - x + 1 - 1 6 - 3 x 2 - x + 1 Now, the first two terms will integrate to natural logarithms and the last term will be a complete the square integral to become the inverse tangent: 1 x 3 + 1 = 1 3 1 x + 1 - 1 6 2 x - 1 x 2 - x + 1 + 1 2 1 x 2 - x + 1 Write each term as an integral ∫ 1 x 3 + 1 d x = 1 3 ∫ 1 x + 1 d x - 1 6 ∫ 2 x - 1 x 2 - x + 1 d x + 1 2 ∫ 1 x 2 - x + 1 d x ∫ 1 x 3 + 1 d x = 1 3 ln ( x + 1 ) - 1 6 ln x 2 - x + 1 + 3 3 tan - 1 2 x - 1 3 + C

Evaluate the integral ∫ sin x cos x cos 2 x + 3 cos x + 2 d x

Let I = ∫ sin x cos x cos 2 x + 3 cos x + 2 d x Put cos x = t - sin x d x = d t sin x d x = - d t ∫ sin x cos x cos 2 x + 3 cos x + 2 d x = ∫ t t 2 + 3 t + 2 d t = ∫ t ( t + 1 ) ( t + 2 ) d t Now, consider t ( t + 1 ) ( t + 2 ) = A t + 1 + B t + 2 ⇒ t = A ( t + 2 ) + B ( t + 1 ) t = A t + 2 A + B t + B t = ( A + B ) t + 2 A + B On comparing coefficients, we get A + B = 1 , 2 A + B = 0 ⇒ A + ( A + B ) = 0 ⇒ A + 1 = 0 A = - 1 - 1 + B = 1 ⇒ B = 2 ∴ t ( t + 1 ) ( t + 2 ) = - 1 t + 1 + 2 t + 2 ⇒ I = ∫ 1 t + 1 d t - 2 ∫ 1 t + 2 d t = [ log ( t + 1 ) ] - 2 [ log ( t + 2 ) ] Hence, I = log cos x + 1 - 2 log cos x + 2 + C

E M B I B E

Intermediate Second Year Mathematics Paper 2B>Calculus>Chapter 6 - Integration>Exercise 6(e)

Andhra Pradesh Board Solutions for Exercise 5: Exercise 6(e)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 5: Exercise 6(e)

Attempt the practice questions from Exercise 5: Exercise 6(e) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2B solutions are prepared by Experienced Embibe Experts.