Integration of Titer+I/square root of titer dx

Answer 1

To integrate the expression Titer + I/√(Titer) dx, we can make use of the substitution method.

Let's substitute u = √(Titer). Then, du = (1/2) * (1/√(Titer)) dTiter.

Rearranging the expression, we have:

√(Titer) dx = 2 du

Now, let's substitute back into the original expression:

(Titer + I/√(Titer)) dx = (u^2 + I/u) * 2 du

Expanding and rearranging, we get:

2u^2 du + 2I du/u

The integral of 2u^2 du is (2/3)u^3 + C.

The integral of 2I du/u is 2I ln|u| + C.

Putting it all together, the final integral is:

(2/3) * (Titer)^(3/2) + 2I ln|√(Titer)| + C

Simplifying further, we have:

(2/3) * (Titer)^(3/2) + 2I ln|(Titer)^(1/2)| + C

Finally, the integral of Titer + I/√(Titer) dx is:

(2/3) * (Titer)^(3/2) + I ln|Titer| + C