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Integration by parts is a technique for performing indefinite integration intudv or definite integration int_a^budv by expanding the differential of a product of functions d(uv) and expressing the original integral in terms of a known integral intvdu. A single integration by parts starts with

 d(uv)=udv+vdu,

(1)

and integrates both sides,

 intd(uv)=uv=intudv+intvdu.

(2)

Rearranging gives

 intudv=uv-intvdu.

(3)

For example, consider the integral intxcosxdx and let

u=x    dv=cosxdx

(4)

du=dx    v=sinx,

(5)

so integration by parts gives

where C is a constant of integration.

The procedure does not always succeed, since some choices of u may lead lớn more complicated integrals phàn nàn the original. For example, consider again the integral intxcosxdx and let

  u=cosx    dv=xdx 
 du=-sinxdx    v=1/2x^2,

(8)

giving

which is more difficult phàn nàn the original (Apostol 1967, pp. 218-219).

Integration by parts may also fail because it leads back lớn the original integral. For example, consider intx^(-1)dx and let

  u=x    dv=x^(-2)dx 
 du=dx    v=-x^(-1),

(11)

then

 intx^(-1)dx=-1-int(-x^(-1))dx+C=intx^(-1)dx+C-1,

(12)

which is same integral as the original (Apostol 1967, p. 219).

The analogous procedure works for definite integration by parts, so

 int_a^budv=[uv]_a^b-int_a^bvdu,

(13)

where [f]_a^b=f(b)-f(a).

Integration by parts can also be applied n times lớn intf^((n))(x)g(x)dx:

  u=g(x)    dv=f^((n))(x)dx 
 du=g^'(x)dx    v=f^((n-1))(x).

(14)

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Therefore,

 intf^((n))(x)g(x)dx=g(x)f^((n-1))(x)-intf^((n-1))(x)g^'(x)dx.

(15)

But

 intf^((n-1))(x)g^'(x)dx=g^'(x)f^((n-2))(x)-intf^((n-2))(x)g^('')(x)dx

(16)

 intf^((n-2))(x)g^('')(x)dx=g^('')(x)f^((n-3))(x)-intf^((n-3))(x)g^((3))(x)dx,

(17)

so

 intf^((n))(x)g(x)dx=g(x)f^((n-1))(x)-g^'(x)f^((n-2))(x) 
 +g^('')(x)f^((n-3))(x)-...+(-1)^nintf(x)g^((n))(x)dx.

(18)

Now consider this in the slightly different khuông intf(x)g(x)dx. Integrate by parts a first time

  u=f(x)    dv=g(x)dx 
 du=f^'(x)dx    v=intg(x)dx,

(19)

so

 intf(x)g(x)dx=f(x)intg(x)dx-int[intg(x)dx]f^'(x)dx.

(20)

Now integrate by parts a second time,

  u=f^'(x)    dv=intg(x)dx 
 du=f^('')(x)dx    v=intintg(x)(dx)^2,

(21)

so

 intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+int[intintg(x)(dx)^2]f^('')(x)dx.

(22)

Repeating a third time,

 intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-int[intintintg(x)(dx)^3]f^(''')(x)dx.

(23)

Therefore, after n applications,

 intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-...+(-1)^(n+1)f^((n))(x)int...int_()_(n+1)g(x)(dx)^(n+1)+(-1)^nint[int...int_()_(n+1)g(x)(dx)^(n+1)]f^((n+1))(x)dx.

(24)

If f^((n+1))(x)=0 (e.g., for an nth degree polynomial), the last term is 0, ví the sum terminates after n terms and

 intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-...+(-1)^(n+1)f^((n))(x)int...int_()_(n+1)g(x)(dx)^(n+1).

(25)


See also

Darboux's Formula, Integral, Integration, Inverse Function Integration, Summation by Parts

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References

Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 12, 1972.Apostol, T. M. "Integration by Parts." §5.9 in Calculus, 2nd ed., Vol. 1: One-Variable Calculus, with an Introduction lớn Linear Algebra. Waltham, MA: Blaisdell, pp. 217-220, 1967.Bronshtein, I. N. and Semendyayev, K. A. Handbook of Mathematics, 3rd ed. New York: Springer-Verlag, p. 269, 1997.

Cite this as:

Weisstein, Eric W. "Integration by Parts." From MathWorld--A Wolfram Web Resource. https://mamnonkidzone.edu.vn/IntegrationbyParts.html

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