calculate the indefinte integral
(x^2+2x+3)/(sqrt of x)
nevermind i find the answer
I assume you divided each numerator term by sqrt x and integrated them separately.
To calculate the indefinite integral of the given expression, divide it into separate fractions and integrate each term individually.
The integral of (x^2)/(sqrt(x)) can be found by applying the power rule for integration:
∫(x^2)/(sqrt(x)) dx = ∫x^(2-1/2) dx = (2/3) * x^(3/2) + C, where C is the constant of integration.
The integral of (2x)/(sqrt(x)) can be solved using u-substitution:
Let u = sqrt(x), then du = (1/2) * x^(-1/2) dx.
Now the integral becomes:
∫(2x)/(sqrt(x)) dx
= 2 ∫(u^2)*(1/2)*(x^(-1/2)) dx
= 2 ∫u^2 du
= (2/3) * u^3 + C
= (2/3) * (sqrt(x))^3 + C
= (2/3) * x^(3/2) + C, where C is the constant of integration.
Finally, the integral of 3/(sqrt(x)) can be found using the power rule for integration again:
∫3/(sqrt(x)) dx = 3 * ∫(1)/(sqrt(x)) dx
= 3 * 2 (√x) + C
= 6√x + C, where C is the constant of integration.
Putting it all together, the indefinite integral of (x^2+2x+3)/(sqrt(x)) is:
(2/3) * x^(3/2) + (2/3) * x^(3/2) + 6√x + C
= (4/3) * x^(3/2) + 6√x + C, where C is the constant of integration.