- #1

Bashyboy

- 1,421

- 5

Hello,

I have tried computing the following integral three times now, and I can not figure out what I am doing wrong.

The integral is [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh[/itex], which wolfram calculates as [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 (-h - \frac{100}{h-10} + c )[/itex].

Here is one sample of my work:

Using integrating by parts, let [itex]u = h^2 - 20h[/itex], which becomes [itex]\frac{du}{dh} = 2h - 20[/itex]. Multiplying both sides by the differential [itex]u[/itex], we get [itex]\frac{du}{dh} dh = (2h-20)dh[/itex]. The definition of the differential of [itex]u[/itex] is [itex]du = \frac{du}{dh} dh[/itex]. Making this substitution, [itex]du = (2h - 20) dh[/itex]. Let [itex]dv = \frac{dh}{(h-10)^2}[/itex]. Integrating both sides, [itex]v = \frac{1}{h-10}[/itex].

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[ (h^2-20h) \left(\frac{1}{h-10} \right) - \int \frac{2h-20}{h-10} dh \right] \implies[/itex]

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int \frac{h-10}{h-10} dh \right] \implies [/itex]

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int dh \right] \implies[/itex]

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2h + c \right] [/itex]

which is not the same...

Can anyone see what I might have done wrong?

I have tried computing the following integral three times now, and I can not figure out what I am doing wrong.

The integral is [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh[/itex], which wolfram calculates as [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 (-h - \frac{100}{h-10} + c )[/itex].

Here is one sample of my work:

Using integrating by parts, let [itex]u = h^2 - 20h[/itex], which becomes [itex]\frac{du}{dh} = 2h - 20[/itex]. Multiplying both sides by the differential [itex]u[/itex], we get [itex]\frac{du}{dh} dh = (2h-20)dh[/itex]. The definition of the differential of [itex]u[/itex] is [itex]du = \frac{du}{dh} dh[/itex]. Making this substitution, [itex]du = (2h - 20) dh[/itex]. Let [itex]dv = \frac{dh}{(h-10)^2}[/itex]. Integrating both sides, [itex]v = \frac{1}{h-10}[/itex].

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[ (h^2-20h) \left(\frac{1}{h-10} \right) - \int \frac{2h-20}{h-10} dh \right] \implies[/itex]

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int \frac{h-10}{h-10} dh \right] \implies [/itex]

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int dh \right] \implies[/itex]

[itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2h + c \right] [/itex]

which is not the same...

Can anyone see what I might have done wrong?

Last edited: