How do you integrate #int (x^2-1)/(x^(3/2))dx#?

Question

Scarlett Allison · Accepted Answer

#int {x^2 - 1}/{x^{3/2}} dx =2/3 x^{3/2} + 2 x^{- 1/2} + C# The integral is solved immediately if we make the previous division of the numerator between the denominator. To do this, we decompose the division into two terms and then simplify the powers of #x#: #int {x^2 - 1}/{x^{3/2}} dx = int (x^2/x^{3/2} - 1/x^{3/2}) dx = int x^{1/2} dx - int x^{- 3/2} dx =# #= {x^{1/2 + 1}}/{1/2 + 1} - {x^{- 3/2 + 1}}/{- 3/2 + 1} + C = 2/3 x^{3/2} + 2 x^{- 1/2} + C#.

Charles Anctil · Answer

undefined To integrate $\int \frac{x^2 - 1}{x^{3/2}} \, dx$, use the method of integration by parts. First, rewrite the integral as $\int x^{2 - 3/2} - x^{-1/2} \, dx$. Then, apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Let $ u = x^{2 - 3/2} $ and $ dv = dx $, so $ du = (2 - 3/2)x^{2 - 3/2 - 1} \, dx = (2 - 3/2)x^{1/2} \, dx $ and $ v = x $.

Substitute these values into the integration by parts formula:

$$ \int x^{2 - 3/2} \, dx - \int x^{-1/2} \, dx = \frac{x^{7/2}}{7/2} - \frac{2x^{1/2}}{1/2} + C $$

$$ = \frac{2}{7}x^{7/2} - 4x^{1/2} + C $$

Therefore, the integral of $ \frac{x^2 - 1}{x^{3/2}} \, dx $ is $ \frac{2}{7}x^{7/2} - 4x^{1/2} + C $, where $ C $ is the constant of integration.