integral ln(x^2-x+2)dx?

Can anyone solve this using integration by parts? I haven't learned the whole integration with i thing so if possible don't use that strategy. Thank you!!

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  • 9 years ago
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    I = ∫ In( x² - x + 2 )dx

    Let :

    . . . . . . . . . . . . . . . . . . . . . 2x - 1

    u = In( x² - x + 2 ) => du = ▬▬▬▬ dx

    . . . . . . . . . . . . . . . . . . . . x² - x + 2

    dv = dx => v = x

    So that :

    I = ∫ In( x² - x + 2 )dx

    . . . . . . . . . . . . . . . x(2x-1)

    = xln(x² - x + 2) - ∫ ▬▬▬▬▬ dx (1)

    . . . . . . . . . . . . . . x² - x + 2

    Let solve :

    . . .2x² - x

    ∫ ▬▬▬▬▬ dx

    . .x² - x + 2

    . . . . . . . . 2x - 5

    = ∫ ( 2 + ▬▬▬▬▬ ) dx

    . . . . . . . x² - x + 2

    . . . . . . . . . . 2x - 1 . . . . . . . . . . 1

    = 2∫ dx + ∫ ▬▬▬▬▬ dx - 4 ∫ ▬▬▬▬▬ dx

    . . . . . . . . . x² - x + 2 . . . . . .x² - x + 2

    = 2x + J + H (2)

    Let solve J :

    . . 2x - 1

    ∫ ▬▬▬▬ dx

    . x² - x + 2

    Let :

    u = x² - x + 2

    du = (2x - 1)dx

    . . du

    ∫ ▬▬▬

    . . u

    = ln|u| + C

    = ln|x² - x + 2| + C (3)

    Let solve H :

    . . . . . . 1

    - 4 ∫ ▬▬▬▬▬ dx

    . . . . x² - x + 2

    . . . . . . . . . 1

    = -4 ∫ ▬▬▬▬▬▬▬ dx

    . . . . .(x - 1/2)² + 7/4

    Let :

    u = x - 1/2

    du = dx

    . . . . . . . . . 1

    = -4 ∫ ▬▬▬▬▬▬ dx

    . . . . . . u² + 7/4

    Let

    u = √7.tanφ / 2( φ ∈ (-π/2 ; π/2 ) ) => φ = arctan( 2u/√7 )

    du = √7 / ( 2cos²φ )

    . . . . . . . . . . dφ

    = -4 ∫ ▬▬▬▬▬▬▬▬▬

    . . . . 7/4( 1 + tan²φ )cos²φ

    . . . .16

    = - ▬▬ ∫ dφ

    . . . .7

    . . . 16

    = - ▬▬φ + C

    . . . .7

    . . . 16

    = - ▬▬arctan( 2u/√7 ) + C

    . . . .7

    . . . 16

    = - ▬▬arctan[ 2(x - 1/2)/√7 ] + C (4)

    . . . .7

    From (1) ; (2) ; (3) ; (4)

    . . . . 16

    => ▬▬▬arctan[ 2(x - 1/2)/√7 ] - ln|x² - x + 2| - 2x + xln(x² - x + 2) + C

    . . . . .7

    Source(s): I'm a Vietnamese student grade 12
  • .
    Lv 4
    9 years ago

    Begin with integration by parts: u=ln(x^2-x+2), dv=dx, du=(2x-1)/(x^2-x+2) dx, v = x.

    The integral becomes uv - integral v du = x ln(x^2-x+2) - integral[(2x^2-x)/(x^2-x+2)]dx.

    Divide the numerator of the integrand by the denominator to simplify this to:

    x ln(x^2-x+2) - integral[2 + (x-4)/(x^2-x+2)]dx.

    The constant term in the integral can easily be integrated right now:

    x ln(x^2-x+2) - 2x - integral[(x-4)/(x^2-x+2)]dx

    Now, complete the square of the denominator of the integrand to convert it to (x-1/2)^2+7/4:

    x ln(x^2-x+2) - 2x - integral[(x-4)/( (x-1/2)^2+7/4 )]dx

    Split the numerator into two terms: (x-1/2)-7/2, and then create two separate integrals:

    x ln(x^2-x+2) - 2x - integral[( (x-1/2) - 7/2 )/((x-1/2)^2+7/4)]dx

    x ln(x^2-x+2) - 2x - integral[(x-1/2)/((x-1/2)^2+7/4)]dx + integral[(7/2)/((x-1/2)^2+7/4)]dx

    Make the substitution u=x-1/2, du=dx:

    x ln(x^2-x+2) - 2x - integral[u/(u^2+7/4)]du + (7/2)integral[1/(u^2+7/4)]du

    The left integral can be solved by substitution: w=u^2, dw=2u du

    x ln(x^2-x+2) - 2x - (1/2)integral[1/(w+7/4)]dw + (7/2)integral[1/(u^2+7/4)]du

    x ln(x^2-x+2) - 2x - (1/2)ln(w+7/4) + (7/2)integral[1/(u^2+7/4)]du

    Remember that w=u^2=(x-1/2)^2:

    x ln(x^2-x+2) - 2x - (1/2)ln((x-1/2)^2+7/4) + (7/2)integral[1/(u^2+7/4)]du

    x ln(x^2-x+2) - 2x - (1/2)ln(x^2-x+2) + (7/2)integral[1/(u^2+7/4)]du

    The logarithmic terms are identical, so we can combine them:

    (x-1/2) ln(x^2-x+2) - 2x + (7/2)integral[1/(u^2+7/4)]du

    The remaining integral could be found in most integral tables, but I'll solve it anyway.

    Do a trigonometric substitution: let u = (sqrt(7)/2) tan z, du = (sqrt(7)/2)sec^2(z) dz

    (x-1/2)ln(x^2-x+2) - 2x + (7/2)(sqrt(7)/2)integral[sec^2(z)/((7/4) tan^2(z)+7/4)]dz

    Since tan^2(z)+1=sec^2(z),

    (x-1/2)ln(x^2-x+2) -2x + (7/2)(sqrt(7)/2)integral[sec^2(z)/((7/4) sec^2(z))]dz

    (x-1/2)ln(x^2-x+2) -2x + (7/2)(sqrt(7)/2)integral[4/7]dz

    Some numerators and denominators cancel:

    (x-1/2) ln(x^2-x+2) - 2x + sqrt(7)integral[1]dz

    (x-1/2) ln(x^2-x+2) - 2x + sqrt(7)z + C

    Since u = (sqrt(7)/2) tan z, z = atan(2u/sqrt(7)). Since u = x-1/2, z = atan((2x-1)/sqrt(7))

    (x-1/2) ln(x^2-x+2) - 2x + sqrt(7) atan((2x-1)/sqrt(7)) + C

    And that's the result!

    (I just checked this using Mathematica, and Mathematica got the same result)

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