16x? In(8x³ +5)6. When finding |dx, which of the following would we let u equal in order to obtain the appropriate8x3 +5u-substitution?a) u = In(8x +5)b) и —1 = 16x²c) u = 8x³ +5d) и %3D 16х? In(&x3+5)e) и %—= 8x3

Given : ∫16x2ln8x3 + 58x3 + 5dx

Answered: 16x² In(8x³ + 5) 6. When finding dx,…

Intermediate Algebra

10th Edition

ISBN:9781285195728

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter8: Conic Sections

Section8.2: More Parabolas And Some Circles

Problem 63.2PS: By expanding (xh)2+(yk)2=r2, we obtain x22hx+h22ky+k2r2=0. When we compare this result to the form...

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16x? In(8x³ +5)
6. When finding |
dx, which of the following would we let u equal in order to obtain the appropriate
8x3 +5
u-substitution?
a) u = In(8x +5)
b) и —
1 = 16x²
c) u = 8x³ +5
d) и %3D 16х? In(&x3+5)
e) и %—
= 8x3

Expert Solution

Step 1 : Analysis

Given :

$\int \frac{16 x^{2} \ln (8 x^{3} + 5)}{8 x^{3} + 5} d x$

Step 2 : Substitution

Let $I = \int \frac{16 x^{2} \ln (8 x^{3} + 5)}{8 x^{3} + 5} d x$

Substituting $u = \ln (8 x^{3} + 5)$ ,

Differentiating both sides with respect to $x$ ,

$\begin{array}{rcl} u & = & \ln (8 x^{3} + 5) \\ \frac{d u}{d x} & = & \frac{d}{d x} \ln (8 x^{3} + 5) \\ \frac{d u}{d x} & = & \frac{1}{(8 x^{3} + 5)} \times \frac{d}{d x} (8 x^{3} + 5) \\ \frac{d u}{d x} & = & \frac{24 x^{2}}{(8 x^{3} + 5)} \\ d u & = & \frac{24 x^{2}}{(8 x^{3} + 5)} d x \\ \frac{(8 x^{3} + 5)}{24 x^{2}} d u & = & d x \end{array}$

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