# 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

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)