Solve: $\int \dfrac{s^{2} + \sqrt{s}}{s^{2}}ds$, step by step please.

Question

I need a step by step example to be able to understand how to solve this problem. I am just now learning the substitution method but I don't know how to apply it here, or if it is applicable at all. Perhaps there is another method I am unaware of and would be glad to be shown such a method. The given interval is [1, sqrt(2)], however, just the indefinite integral would suffice.

Answer 1

HINT: $$\int \frac{s^2+\sqrt s}{s^2}\ ds=\int \left(\frac{s^2}{s^2}+\frac{\sqrt s}{s^2}\right)\ ds=\int \left(s^{0}+s^{-3/2}\right)\ ds $$ Use can now use $\large \int x^n\ dx=\frac{x^{n+1}}{n+1}$

Answer 2

$\int \dfrac{s^{2} + \sqrt{s}}{s^{2}}ds=\int (1+ s^{-3/2})ds$

Answer 3

HINT:

$$\int\frac{s^2+\sqrt{s}}{s^2}\space\text{d}s=$$

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Substitute $u=\sqrt{s}$ and $\text{d}u=\frac{1}{2\sqrt{s}}\space\text{d}s$:

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$$2\int\frac{u^3+1}{u^2}\space\text{d}u=$$ $$2\int\left(\frac{1}{u^2}+u\right)\space\text{d}u=$$ $$2\int\frac{1}{u^2}\space\text{d}u+2\int u\space\text{d}u=$$ $$2\int u^{-2}\space\text{d}u+2\int u\space\text{d}u$$