Find the integral ∫ 1 x 2 - x d x

Let I = ∫ 1 x 2 - x d x Consider, 1 - x x - 2 = a 0 x + a 1 x − 2 ⇒ 1 = − a 0 ( x − 2 ) − a 1 x ⇒ 1 = − a 0 x + 2 a 0 − a 1 x ⇒ 1 = x − a 0 − a 1 + 2 a 0 On comparing the coefficients of like powers of x , we get: 2 a 0 = 1 , − a 0 − a 1 = 0 On solving the equations, we get: a 0 = 1 2 , a 1 = − 1 2 ∴ I = ∫ 1 2 x − 1 2 ( x − 2 ) d x = ∫ 1 2 x d x − ∫ 1 2 ( x − 2 ) d x = ln x 2 − ln x - 2 2 + c = 1 2 ln x x - 2 + c ∵ ln p - ln q = ln p q

Find the integral ∫ 2 x - 5 x + 2 x - 1 d x

Let I = ∫ 2 x - 5 x + 2 x - 1 d x Consider, 2 x - 5 x + 2 x - 1 = a 0 x + 2 + a 1 x − 1 ⇒ 2 x − 5 = a 0 ( x − 1 ) + a 1 ( x + 2 ) ⇒ 2 x − 5 = a 0 x − a 0 + a 1 x + 2 a 1 ⇒ 2 x − 5 = x a 0 + a 1 − a 0 + 2 a 1 On comparing the coefficients of like terms of x , we get: 2 = a 0 + a 1 and - 5 = − a 0 + 2 a 1 On solving the above equations, we get: a 0 = 3 , a 1 = − 1 Thus, I = ∫ 3 x + 2 − 1 x − 1 d x ⇒ I = ∫ 3 x + 2 d x − ∫ 1 x − 1 d x ⇒ I = 3 ln x + 2 − ln x - 1 + c

Find the integral ∫ 2 4 x - 9 2 x + 1 x - 5 d x

Let I = ∫ 2 4 x - 9 2 x + 1 x - 5 d x Consider, 8 x − 18 ( x − 5 ) ( 2 x + 1 ) = a 0 2 x + 1 + a 1 x − 5 ⇒ 8 x − 18 = a 0 ( x − 5 ) + a 1 ( 2 x + 1 ) ⇒ 8 x − 18 = a 0 x − 5 a 0 + 2 a 1 x + a 1 ⇒ 8 x − 18 = x a 0 + 2 a 1 − 5 a 0 + a 1 On comparing the coefficients of like powers of x , we get: 8 = a 0 + 2 a 1 and − 5 a 0 + a 1 = - 18 On solving the above equations, we get: a 0 = 4 , a 1 = 2 Thus, I = ∫ 4 2 x + 1 + 2 x − 5 d x ⇒ I = ∫ 4 2 x + 1 d x + ∫ 2 x − 5 d x ⇒ I = 4 ln 2 x + 1 2 + 2 ln x - 5 + c ⇒ I = 2 ln 2 x + 1 + 2 ln x - 5 + c ⇒ I = 2 ln 2 x + 1 + ln x - 5 + c ⇒ I = 2 ln 2 x + 1 × x - 5 + c ∵ ln m · n = ln m + ln n ⇒ I = 2 ln 2 x 2 - 10 x + x - 5 + c ⇒ I = 2 ln 2 x 2 - 9 x - 5 + c

Find the integral ∫ x 2 + 2 x - 5 x - 3 x 2 + 1 d x

Let I = ∫ x 2 + 2 x - 5 x - 3 x 2 + 1 d x Consider, x 2 + 2 x - 5 x - 3 x 2 + 1 = a 0 x − 3 + a 2 x + a 1 x 2 + 1 ⇒ x 2 + 2 x − 5 = a 0 x 2 + 1 + a 2 x + a 1 ( x − 3 ) ⇒ x 2 + 2 x − 5 = a 0 x 2 + a 0 + a 2 x 2 - 3 a 2 x + a 1 x - 3 a 1 ⇒ x 2 + 2 x − 5 = x 2 a 0 + a 2 + x - 3 a 2 + a 1 - 3 a 1 + a 0 On comparing the coefficients of like powers of x , we get: 1 = a 0 + a 2 , - 3 a 2 + a 1 = 2 , - 3 a 1 + a 0 = - 5 On solving the above equations, we get: a 0 = 1 , a 2 = 0 , a 1 = 2 Thus, I = ∫ 1 x − 3 + 2 x 2 + 1 d x ⇒ I = ∫ 1 x − 3 d x + ∫ 2 x 2 + 1 d x ⇒ I = ln x - 3 + 2 tan - 1 x + c

Find the integral ∫ 6 x 2 - 19 x + 20 x + 1 x - 2 2 d x

Let I = ∫ 6 x 2 - 19 x + 20 x + 1 x - 2 2 d x Consider, 6 x 2 - 19 x + 20 x + 1 x - 2 2 = a 0 x + 1 + a 1 x − 2 + a 2 ( x − 2 ) 2 ⇒ 6 x 2 − 19 x + 20 = a 0 ( x − 2 ) 2 + a 1 ( x + 1 ) ( x − 2 ) + a 2 ( x + 1 ) ⇒ 6 x 2 − 19 x + 20 = a 0 x 2 + 4 a 0 - 4 a 0 x + a 1 ( x 2 - x - 2 ) + a 2 x + a 2 ⇒ 6 x 2 − 19 x + 20 = a 0 x 2 + 4 a 0 - 4 a 0 x + a 1 x 2 - a 1 x - 2 a 1 + a 2 x + a 2 ⇒ 6 x 2 − 19 x + 20 = x 2 a 0 + a 1 + x - 4 a 0 - a 1 + a 2 - 2 a 1 + a 2 + 4 a 0 On comparing the coefficients of like powers of x , we get: 6 = a 0 + a 1 , - 19 = - 4 a 0 - a 1 + a 2 , 20 = - 2 a 1 + a 2 + 4 a 0 On solving the above equations, we get: a 0 = 5 , a 1 = 1 , a 2 = 2 Thus, I = ∫ 5 x + 1 + 1 x − 2 + 2 ( x − 2 ) 2 d x ⇒ I = ∫ 5 x + 1 d x + ∫ 1 x − 2 d x + ∫ 2 ( x − 2 ) 2 d x ⇒ I = 5 ln x + 1 + ln x - 2 - 2 x - 2 + c

Find the integral ∫ 2 x 2 + 4 x - 21 x - 1 x + 4 d x

Let I = ∫ 2 x 2 + 4 x - 21 x - 1 x + 4 d x Consider, 2 x 2 + 4 x - 21 x - 1 x + 4 = 2 + − 2 x − 13 x 2 + 3 x − 4 [applying long division] Now, consider − 2 x − 13 x 2 + 3 x − 4 = a 0 x − 1 + a 1 x + 4 ⇒ − 2 x − 13 = a 0 ( x + 4 ) + a 1 ( x − 1 ) ⇒ − 2 x − 13 = a 0 x + 4 a 0 + a 1 x − a 1 ⇒ − 2 x − 13 = x a 0 + a 1 + 4 a 0 − a 1 On comparing the coefficients of like powers of x , we get: - 2 = a 0 + a 1 , - 13 = 4 a 0 − a 1 On solving the above equation , we get: a 0 = − 3 , a 1 = 1 Thus, I = ∫ 2 + - 3 x − 1 + 1 x + 4 d x ⇒ I = ∫ 2 d x + ∫ - 3 x − 1 d x + ∫ 1 x + 4 d x ⇒ I = 2 x - 3 ln x - 1 + ln x + 4 + c

Evaluate the integral ∫ 1 2 5 x + 13 x + 2 x + 3 d x

Let I = ∫ 1 2 5 x + 13 ( x + 2 ) ( x + 3 ) d x Consider, 5 x + 13 ( x + 2 ) ( x + 3 ) = a 0 x + 2 + a 1 x + 3 ⇒ 5 x + 13 = a 0 ( x + 3 ) + a 1 ( x + 2 ) ⇒ 5 x + 13 = a 0 x + 3 a 0 + a 1 x + 2 a 1 ⇒ 5 x + 13 = x a 0 + a 1 + 3 a 0 + 2 a 1 On comparing the coefficients of like powers of x , we get: 5 = a 0 + a 1 , 13 = 3 a 0 + 2 a 1 On solving these, we get: a 0 = 3 , a 1 = 2 Thus, I = ∫ 1 2 3 x + 2 + 2 x + 3 d x ⇒ I = 3 ln x + 2 + 2 ln x + 3 1 2 ⇒ I = 3 ln 2 + 2 + 2 ln 2 + 3 - 3 ln 1 + 2 + 2 ln 1 + 3 ⇒ I = 3 ln 4 + 2 ln 5 - 3 ln 3 + 2 ln 4 ⇒ I = ln 4 3 + ln 5 2 - ln 3 3 + ln 4 2 ∵ ln p q = q ln p ⇒ I = ln 64 + ln 25 - ln 27 + ln 16 ⇒ I = ln 64 × 25 - ln 27 × 16 ∵ ln p × q = ln p + ln q ⇒ I = ln 1600 - ln 432 ⇒ I = ln 1600 432 ∵ ln p q = ln p - ln q ⇒ I = ln 100 27

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 7

MEDIUM

AS and A Level

IMPORTANT

Earn 100

The diagram shows part of the curve $y = \frac{\ln x}{\sqrt{x}} .$ Using the substitution $u = \ln x,$ find the exact value of the volume of the solid formed when the shaded region is rotated completely about the $x$ -axis.

Important Questions on Further Calculus

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8D>Q 8

The diagram shows part of the curve $y = 3 e^{\cos x} \sin x .$ The curve has rotational symmetry of order $2$ about the origin. Using the substitution $u = \cos x,$ find the exact value of the total area bounded by the curve and the $x$ -axis between $x = - π$ and $x = π .$