What is the integral from 2 to 1 of (4+u^2)/(u^3)?

Answer 1

You said the integral of 2 to 1, but I'm going to assume you meant 1 to 2.

Integral (1 to 2, (4 + u^2)/u^3) du

To solve this, let's split this up into two fractions.

(4 + u^2)/u^3 = 4/(u^3) + (u^2)/(u^3)

Now, we can reduce the second fraction.

(4 + u^2)/u^3 = 4/(u^3) + 1/u

Also, note that the u^3 can be brought up to the numerator as a negative exponent.

(4 + u^2)/u^3 = 4u^(-3) + 1/u

So now, we integrate this.

Integral [4u^(-3) + 1/u] du

Let's split this up into two integrals.

Integral (4u^(-3))du + Integral (1/u)du

For the first integral, let's pull out that constant 4.

4 * Integral (u^(-3))du + Integral (1/u) du

Now, we integrate normally. Note that we just use the reverse power rule for the first one, and for the second one, the integral is just ln|u|.

4( u^(-2)/(-2) ) + ln|u|

Simplifying this slightly, we have

-2u^(-2) + ln|u|
(-2)/(u^2) + ln|u|

Now, we evaluate this from 1 to 2.

[(-2)/(2^2) + ln|2|] - [(1)/(1^2) + ln|1|]

[(-1/2) + ln(2)] - [1 + 0]

(-1/2) + ln(2) - 1

Merging the two numbers, we have

ln(2) - 3/2