# 21. Integration (III) - task solutions (1)

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MATH1110 Lecture tasks - Worked solutions Integration (III) Task 1. We will use integration by parts, where u ( x ) = x and v 0 ( x ) = e 2 x . We have u 0 ( x ) = 1 and v ( x ) = 1 2 e 2 x (using integration by substitution to reverse the Chain rule), and so Z 1 0 xe 2 x dx = 1 2 xe 2 x 1 0 - Z 1 0 1 2 e 2 x dx = 1 2 xe 2 x 1 0 - 1 4 e 2 x 1 0 (by again using substitution if needed) = 1 2 e - 0 - 1 4 e - 1 4 = e + 1 4 . Task 2. Let I ( x ) = R e x sin( x ) dx . We will use integration by parts, where u ( x ) = sin( x ) and v 0 ( x ) = e x . We have u 0 ( x ) = cos( x ) and v ( x ) = e x , and so I ( x ) = Z e x sin( x ) dx = e x sin( x ) - Z e x cos( x ) dx. To evaluate the integral R e x cos( x ) dx we will again use integration by parts, where u ( x ) = cos( x ) and v 0 ( x ) = e x . We have u 0 ( x ) = - sin( x ) and v ( x ) = e x , and so Z e x cos( x ) dx = e x cos( x ) - Z e x ( - sin( x )) dx = e x cos( x ) + Z e x sin( x ) dx. Combining this with the result for I ( x ) allows us to conclude I ( x ) = Z e x sin( x ) dx = e x sin( x ) - Z e x cos( x ) dx = e x sin( x ) - e x cos( x ) + Z e x sin( x ) dx = e x sin( x ) - e x cos( x ) - I ( x ) = 2 I ( x ) = e x sin( x ) - e x cos( x ) = I ( x ) = 1 2 e x sin( x ) - 1 2 e x cos( x ) . Page 1
MATH1110 Lecture tasks - Worked solutions Integration (III) Task 3. Note that the denominator x 2 + 2 x may be factorised as x ( x + 2), and so using the method of partial fractions we begin by expressing this fraction in the form 1 x ( x + 2) A x + B x + 2 = A ( x + 2) + Bx x ( x + 2) = ( A + B ) x + 2 A x ( x + 2) . This leads to the simultaneous equations A + B = 0 and 2 A = 1, from which we can conclude A = 1 / 2 and B = - 1 / 2. Thus, we obtain 1 x ( x + 2) = 1 / 2 x - 1 / 2 x + 2 , allowing us to evaluate the indefinite integral Z 1 x 2 + 2 x dx = Z 1 / 2 x - 1 / 2 x + 2 dx = 1 2 ln | x | - 1 2 ln | x + 2 | + C. Task 4. Note that the denominator x 2 + 2 x + 1 may be factorised as ( x + 1) 2 , and so using the method of partial fractions we begin by expressing this fraction in the form 1 ( x + 1) 2 A x + 1 + B ( x + 1) 2 = A ( x + 1) + B ( x + 1) 2 = ( Ax + ( A + B ) ( x + 1) 2 . This leads to the simultaneous equations A = 0 and A + B = 1, from which we can conclude A = 0 and B = 1. Thus, we obtain 1 ( x + 1) 2 = 0 x + 1 + 1 ( x + 1) 2 , (why did we even bother doing this?) allowing us to evaluate the indefinite integral Z 1 x 2 + 2 x + 1 dx = Z 1 ( x + 1) 2 dx = - 1 x + 1 + C. (using substitution if needed) Task 5. Note that the denominator cannot be factorised over the real numbers, and we have learned from the last example that partial fractions is a bit pointless when there is only one factor in the denominator and the numerator is a 1. Instead, we complete the square on the denominator to rewrite x 2 + 2 x + 2 as ( x + 1) 2 + 1. Using integration by substitution (if needed) allows us to evaluate the indefinite integral Z 1 x 2 + 2 x + 2 dx = Z 1 ( x + 1) 2 + 1 dx = tan - 1 ( x + 1) + C. Page 2