INDEFINITE INTEGRAL OF RATIONAL FUNCTIONS
Problem 1 :
2 1 3 x 2 - 1 dx
Solution :
Given , 2 1 3 x 2 - 1 dx = 2 1 3x -2 - 1 dx = 2 1 3x -2 dx - 2 1 1 dx = 3x -1 - 1 - x 2 1 = - 3 x - x 2 1 = - 3 2 - 2 - - 3 1 - 1 = - 3 2 - 2 + 3 + 1 = - 3 2 - 2 + 4 = - 3 2 + 2 = - 3 + 4 2 = 1 2
Problem 2 :
-1 -2 u - 1 u 2 du
Solution :
Given , -1 -2 u - 1 u 2 du = -1 -2 u - u -2 du = -1 -2 u du - -1 -2 u -2 du = u 2 2 - u -1 - 1 -1 -2 = u 2 2 + 1 u -1 -2 = ( - 1 ) 2 2 + 1 - 1 - ( - 2 ) 2 2 + 1 - 2 = 1 2 - 1 1 - 4 2 + 1 2 = 1 2 - 1 - 2 + 1 2 = 1 - 2 - 4 + 1 2 = - 4 2 = - 2
Problem 3 :
4 1 u - 2 u du
Solution :
Given , 4 1 u - 2 u du = 4 1 u u - 2 u du = 4 1 u - 2 u du = 4 1 u 1 2 - 2u - 1 2 du = u 1 2 +1 1 2 + 1 - 2 × u - 1 2 + 1 - 1 2 + 1 4 1 = u 3 2 3 2 - 2 × u 1 2 1 2 4 1 = ( 4 ) 3 2 3 2 - 2 × ( 4 ) 1 2 1 2 4 1 - ( 1 ) 3 2 3 2 - 2 × ( 1 ) 1 2 1 2 4 1 = ( 2 ) 3 3 2 - 2 × 2 1 2 - 1 3 2 - 2 × 1 1 2 = 2 3 × 8 - 8 - 2 3 - 4 = 16 3 - 8 - 2 3 + 4 = 14 3 - 4 = 14 - 12 3 = 2 3
Problem 4 :
8 1 2 x dx
Solution :
Given , 8 1 2 x dx = 8 1 2 ( x ) - 1 2 dx = 2 x - 1 2 + 1 - 1 2 + 1 8 1 = 2 x - 1 + 2 2 -1 + 2 2 8 1 = 2 x 1 2 1 2 8 1 = 2 2x 1 2 8 1 = 2 2 x 8 1 = 2 2 8 - 1 = 2 2 2 2 - 1 = 4 × 2 - 2 2 = 2 4 - 2
Problem 5 :
1 0 x - x 3 dx
Solution :
Given , 1 0 x - x 3 dx = 1 3 1 0 x - x dx = 1 3 1 0 x - x 1 2 dx = 1 3 1 0 x dx - 1 0 x 1 2 dx = 1 3 x 2 2 1 0 - x 1 2 + 1 1 2 + 1 1 0 = 1 3 x 2 2 1 0 - x 1 + 2 2 1 + 2 2 1 0 = 1 3 x 2 2 1 0 - x 3 2 3 2 1 0 = 1 3 x 2 2 1 0 - x 3 2 × 2 3 1 0 = x 2 6 1 0 - x 3 2 × 2 9 1 0 = 1 2 6 - 0 2 6 - 1 3 2 × 2 9 - 0 3 2 × 2 9 = 1 6 - 2 9 = 3 - 4 18 = - 1 18
Problem 6 :
-1 -8 x - x 2 2∛x dx
Solution :
Problem 7 :
𝜋 4 0 1 - sin 2 𝜃 cos 2 𝜃 d𝜃
Solution :
Given , 𝜋 4 0 1 - sin 2 𝜃 cos 2 𝜃 d𝜃 = 𝜋 4 0 cos 2 𝜃 cos 2 𝜃 d𝜃 = 𝜋 4 0 1 d𝜃 = [ 𝜃 ] 𝜋 4 0 = 𝜋 4 - 0 = 𝜋 4
Problem 8 :
e 1 2x + 1 x dx
Solution :
Given , e 1 2x + 1 x dx = e 1 2x dx + e 1 1 x dx = 2x 2 2 e 1 + x x 2 e 1 = x 2 e 1 + 1 x e 1 = e 2 - 1 2 + [ In | e | - In | 1 | ] = e 2 - 1 + 1 = e 2
Problem 9 :
5 1 x + 1 x dx
Solution :
Given , 5 1 x + 1 x dx = 5 1 x x dx + 5 1 1 x dx = = 5 1 1 dx + 5 1 1 x dx = [ x ] 5 1 + In | x | 5 1 = [ 5 - 1 ] + [ In | 5 | - In | 1 | ] = 4 + [ In | 5 | - In | 1 | ] = 4 + In 5
Problem 10 :
2e e cos x - 1 x dx
Solution :
Given , 2e e cos x - 1 x dx = cos x - 1 x dx = cos x dx - 1 x dx = [ sin x - In | x | ] 2e e = [ ( sin ( 2e ) - In | 2e | ) - ( sin e - In | e | ] = sin ( 2e ) - In | 2e | - sin e + In | e | = sin 2e - sin e + In | e | | 2e | = sin 2e - sin e + In 1 2 = sin 2e - sin e - In 2