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GeezerVison is showing mathworld.wolfram.com — Integration by Parts -- from Wolfram MathWorld

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Integration by Parts

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

intxcosxdx=xsinx-intsinxdx
(6)
=xsinx+cosx+C,
(7)

where C is a constant of integration.

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

u=cosx dv=xdx du=-sinxdx v=1/2x^2,
(8)

giving

intxcosxdx=1/2x^2cosx-1/2intx^2(-sinx)dx
(9)
=1/2x^2cosx+1/2intx^2sinxdx,
(10)

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

Integration by parts may also fail because it leads back to 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 to intf^((n))(x)g(x)dx:

u=g(x) dv=f^((n))(x)dx du=g^'(x)dx v=f^((n-1))(x).
(14)

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 form 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=∬g(x)(dx)^2,
(21)

so

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)∬g(x)(dx)^2+int[∬g(x)(dx)^2]f^('')(x)dx.
(22)

Repeating a third time,

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)∬g(x)(dx)^2+f^('')(x)∭g(x)(dx)^3-int[∭g(x)(dx)^3]f^(''')(x)dx.
(23)

Therefore, after n applications,

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)∬g(x)(dx)^2+f^('')(x)∭g(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, so the sum terminates after n terms and

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)∬g(x)(dx)^2+f^('')(x)∭g(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 to 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.

Referenced on Wolfram|Alpha

Integration by Parts

Cite this as:

Weisstein, Eric W. "Integration by Parts." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/IntegrationbyParts.html

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