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    Integration by Parts

    Uploaded by

    Rodziah Nasir
    The document discusses integration by parts and partial fractions methods. 1) Integration by parts involves setting one function as 'u' and the derivative of the other as 'dv'. The integral is then expressed as uv - integral of v du. This may need to be repeated if the integral remains in terms of two functions. 2) Partial fractions decomposes rational functions into simpler fractional components involving linear, repeated and quadratic terms. This allows each component to be integrated separately. The method involves setting the rational function equal to the partial fractions and solving for the unknown coefficients. 3) Examples demonstrate applying both techniques to evaluate definite integrals involving trigonometric, logarithmic, exponential and rational functions.

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    Attribution Non-Commercial (BY-NC)
    Download as DOCX, PDF, TXT or read online from Scribd

    Integration by Parts

    Uploaded by

    Rodziah Nasir
    441 views3 pages
    The document discusses integration by parts and partial fractions methods. 1) Integration by parts involves setting one function as 'u' and the derivative of the other as 'dv'. The integral is then expressed as uv - integral of v du. This may need to be repeated if the integral remains in terms of two functions. 2) Partial fractions decomposes rational functions into simpler fractional components involving linear, repeated and quadratic terms. This allows each component to be integrated separately. The method involves setting the rational function equal to the partial fractions and solving for the unknown coefficients. 3) Examples demonstrate applying both techniques to evaluate definite integrals involving trigonometric, logarithmic, exponential and rational functions.

    Original Title

    Integration by Parts (6)

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    © Attribution Non-Commercial (BY-NC)

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    The document discusses integration by parts and partial fractions methods. 1) Integration by parts involves setting one function as 'u' and the derivative of the other as 'dv'. The integral is then expressed as uv - integral of v du. This may need to be repeated if the integral remains in terms of two functions. 2) Partial fractions decomposes rational functions into simpler fractional components involving linear, repeated and quadratic terms. This allows each component to be integrated separately. The method involves setting the rational function equal to the partial fractions and solving for the unknown coefficients. 3) Examples demonstrate applying both techniques to evaluate definite integrals involving trigonometric, logarithmic, exponential and rational functions.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    441 views3 pages

    Integration by Parts

    Uploaded by

    Rodziah Nasir
    The document discusses integration by parts and partial fractions methods. 1) Integration by parts involves setting one function as 'u' and the derivative of the other as 'dv'. The integral is then expressed as uv - integral of v du. This may need to be repeated if the integral remains in terms of two functions. 2) Partial fractions decomposes rational functions into simpler fractional components involving linear, repeated and quadratic terms. This allows each component to be integrated separately. The method involves setting the rational function equal to the partial fractions and solving for the unknown coefficients. 3) Examples demonstrate applying both techniques to evaluate definite integrals involving trigonometric, logarithmic, exponential and rational functions.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as DOCX, PDF, TXT or read online from Scribd
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    INTEGRATION BY PARTS

    1)

    Solve the integral method Solution:

    using integration by parts

    Step 2: = =

    ( )

    ( ) *9 +

    Step 1: Determine u and dv , their derivative and integration.

    = =

    u=? du = ? u = x2 du = 2x dx

    = ? v=?

    3)

    Solve the integral

    using integration by parts method

    Step 1: = sin x dx v = - cos x

    u = ex du = ex dx

    = sin x dx v = - cos x

    Step 2: Arrange the equation into correct order as below = Step 3: If still in form of two functions, repeat the

    Step 2: = = -

    dx

    Step 3:

    Step 1 again. = u1 = 2x du1 = 2 dx = cos x dx v1 = sin x Step 4: Arrange the equation properly and solve the integration part. = = = [ EXERCISE ] Step 4: = =

    u1 = du1 =

    = cos x dx v1 = sin x

    2I
    ; :

    Solve the following integrals using integration by parts: a)

    6 3

    2)

    Solve the integral method Step 1: u = ln x du = dx

    using integration by parts b) c) = x dx v=


    2

    5
    4

    (4)

    31

    INTEGRATION USING PARTIAL FRACTION 2) There are 4 types of partial fraction: o Linear function o Repeated factor function o Quadratic term function o Improper fraction
    : ; :

    Integrate

    ; :

    ; :

    using partial fraction

    Solution: Step 1: Step 2:


    ; : ; :

    5 5 4 4 3

    1)

    Integrate

    using partial fraction Step 3:

    Solution: Step 1: Separate into partial fraction

    When x = 1, (-1)2 + 2 5 = A(1) + B(0) + C(0) A = -2 Compare power of x, x 1=A+B 1 = (-2) + B


    2

    3
    Step 2: Multiply the equation 3 3 Step 3: Solve for the unknowns x=0 x=1 x = -3 When x = 0, 0 + 1 = A (0 + 0 3) + 0 + 0 A= When x = 1, ( B= When x = -3, 3 C= (
    6

    3 3

    B=3 x -2 = 4A + 3B + C -2 = 4(-2) + 3(3) + C C = -3

    Step 4:

    ; :

    ; :

    3 ) 3 3 3) Integrate

    using partial fraction

    ( )

    Step 1:

    : :

    Step 2: x2 = A(x2 + 1) + (Bx + C) (x 2) x2 = A(x2 + 1) + B(x2 2x) + C(x 2)

    Step 4: Complete the partial fraction & integrate one by one.

    : ; :

    Step 3:

    ; 6

    When x = 2, 4 = A(5) + B(0) + C(0) A=


    4

    Compare power of x, x2 1 = A + B 1= B=
    4

    +B

    x 0 = -2B + C = ( )
    5

    C=
    32

    Step 4:

    4 5
    4

    4 ;

    : :

    : ;

    4)

    Integrate

    :6 ;

    ; 8

    using partial fraction

    Step 1: 3

    x-2
    4
    4

    5
    3 3

    6 ddd i -18

    x 3x

    -2x + 6x2 -2x3 + 6x2


    4

    6 3

    Step 2: 18 = A(x2-3x) + B(x-3) + C(x2) Step 3: When x = 0, B = -6 Compare power of x, x2 0 = A + C A = -2 Step 4: When x = 3, C=2

    6 ;

    +
    3

    EXERCISE Solve the following integrals using partial fraction: a)

    ;7 ; ; ; ;;

    b)

    c)

    :; ; : ; : :

    d)

    33

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