Let the functions u=u(x) and v=v(x) have continuous derivatives, then the derivative formula of the product of the two functions is

(ultraviolet)' = u' v+ ultraviolet'

Phase shift uv'=(uv)'-u'v

Find the indefinite integral on both sides of this equation and get

∫uv'dx=uv-∫u'vdx ( 1)

The formula (1) is called the partial integral formula. If it is difficult to find ∫uv'dx, but easy to find ∫u'vdx, then the partial integral formula can play a role.

For simplicity, the formula (1) can also be written in the following form.

∫udv=uv-∫vdu

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Now through the column to illustrate.

Column 1. Find ∫xlnxdx

Solution: let u = lnx and dv = dx, then

∫xlnxdx=∫lnxd(x^2/2)

=x^2/2lnx-∫x^2/2d(lnx)

=x^2/2lnx- 1/2∫xdx

=x^2/2lnx-x^2/4+C

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Column two. ∫arccosxdx

Solution: let u = arccosx and dv = dx, then

∫arccosxdx = xarccosx-∫xd(arccosx)

=xarccosx+∫x/√( 1-x^2)dx

=xarccosx- 1/2∫ 1/( 1-x^2)^ 1/2d( 1-x^2)

=xarccosx-√( 1-x^2)+C

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Summary 1: After being skillfully used in partial integration, it is no longer necessary to write out which division chooses U and which division chooses dv. As long as the integrand expression is in the form of φ(x)dv(x), the partial integral formula can be used.

Question 3. Find ∫ x 2sin 2xdx

Solution: power consumption reduction is a priority, because sin2x =1/2 (1-cos2x), so

∫x^2sin^2xdx= 1/2∫x^2( 1-cos2x)dx= 1/6x^3- 1/4∫x^2dsin(2x).

Continuous use of parts integration

∫x^2sin^2xdx= 1/6x^3- 1/4x^2sin2x+ 1/2∫xsin2xdx= 1/6x^3- 1/4x^2sin2x- 1/4∫xdcos2x

= 1/6x^3- 1/4x^2sin2x- 1/4xcos2x+ 1/8sin2x+c

Column four. ∫ x 2e xdx

Solution: let u = x 2, dv = e xdx = d (e x), then

∫x^2e^xdx=∫x^2d(e^x)=x^2e^x-∫e^xd(x^2)-2∫xe^xdx

Here, ∫ Xe XDX is easier to integrate than ∫ X 2e XDX, because the power of x in the integrand is reduced once, so it is enough to use partial integration for ∫ Xe XDX, so

∫x^2e^xdx=x^2e^x-2∫xe^xdx=x^2e^x-2∫xd(e^x)

=x^2e^x-2(xe^x-e^x)+C

=e^x(x^2-2x+2)+C

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Summary 2: The above two columns show that if the integrand is the product of a power function and a sine (cosine) function or a power function and an exponential function, we can consider using partial integration and setting the power function as u, so that the power of the power function can be reduced once by partial integration until the answer is found, and the power exponent assumed here is a positive integer.

Question 5. Find ∫xarctanxdx

Solution: ∫ xarctanxdx =1/2 ∫ arctanxd (x2)

=x^2/2arctanx- 1/2∫x^2/( 1+x^2)dx

=x^2/2arctanx- 1/2∫( 1+x^2- 1)/( 1+x^2)dx

=x^2/2arctanx- 1/2∫[ 1- 1/( 1+x^2)]dx

=x^2/2arctanx- 1/2(x-arctanx)+c

= 1/2(x^2+ 1)arctanx- 1/2x+c

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Summary 3: If the integrand is the product of a power function and a logarithmic function or a power function and an inverse trigonometric function, we can consider partial integration and set the logarithmic function or the inverse trigonometric function as u. 。

Column six. e xsinxdx

Solution: ∫ e xsin xdx = ∫ sinxd (e x) = e xsin x-∫ e xcos xdx

The integral at the right end of the equation is of the same type as the integral at the left end of the equation. Take the integral on the right and then use the partial integral, and you will get

∫e^xsinxdx=e^xsinx-∫cosxd(e^x)

=e^xsinx-e^xcosx-∫e^xsinxdx

Since the third term at the right end of the above formula is the required integral ∫ e xsinxdx, move it to the left end of the equal sign, and then divide the two ends by 2, and you get it.

∫e^xsinxdx= 1/2e^x(sinx-cosx)+c

Since the right end of the above formula no longer contains the integral term, an arbitrary constant c must be added.

Through partial integration, the corresponding problems are summarized. If friends can't understand them well, we have the following table, which can be more conducive to your understanding.

(1) must first be written in the form of ∫udv (or ∫uv'dx).

(2) The partial integral is applied for many times, and the integral of each part is simplified once until it is finally solved.

(3) Partial integration can sometimes derive the equation of ∫f(x)dx and then solve it.

(4) Recursive formulas can sometimes be derived from partial integrals.

When studying indefinite integral of partial integral in college advanced mathematics, you can usually master the first three kinds, even the last question in the exam can't escape this range. For students who take the postgraduate entrance examination (only logarithmic one), you need to do more questions and understand the recursive formula derived from partial integrals.