Unformatted text preview: Example
Evaluate the indeﬁnite integral
√ 6x 3
5x4 + 12 dx. Solution: This problem is made diﬃcult because of the square root of a function of x in the
denominator. However, a simpliﬁcation occurs when we make an appropriate substitution.
Let u = u(x) denote the function of x under the square root, i.e.,
u = 5x4 + 12, (1) and note that the diﬀerential du satisﬁes
du = 20x3 dx. (2) The
√ reason that u is a good substitution in this problem is that, after replacing the expression
√
5x4 + 12 in the integrand by u, the remaining xdependent factor, x3 dx, can be seen to
be a constant multiple of du given above. That is, we can divide through by 20 in (2) to
obtain
1
(3)
x3 dx = du,
20
and obtain a simpliﬁcation of the original integral occurs when the substitutions (1) and (3)
are made; i.e.,
6x3
dx = 6
3x4 + 12 1 x3 dx
+ 12
1
1
=6 √·
du
u 20
3 −1/2
u
du,
=
10
an indeﬁnite integral in the new variable of integration u, and one for which the solution is
easily found. Thus
√ √ √ 3x4 6x3
3 u1/2
dx =
+ C.
10 1/2
3x4 + 12 Now replacing u by 5x4 + 12 in the solution given above, it follows that
6x3
3 (5x4 + 12)1/2
dx =
+C
10
1/2
3x4 + 12
3√ 4
=
5x + 12 + C,
5
so that the ﬁnal result is entirely in terms of x, the original variable of integration.
√ ...
View
Full
Document
This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus

Click to edit the document details