Solve integration using u substitution of (x+1)sqrt(2-x)dx
integrate (x+1)sqrt(2-x)dx
u = (sqrt(2 - x))
u^2 = 2 - x
x = 2 - u^2
- 2u du = dx
I = integral sign
I (2 - u^2 + 1)u -2u du
I (3 - u^2)-2u^2 du
I (-6u^2 + 2u^4) du
-6 I u^2 du + 2 I u^4 du
2 I u^4 du - 6 I u^2 du
2 (1/5 u^5) - 6 (1/3 u^3)
2/5 u^5 - 2u^3
substitute back in for u = (sqrt(2-x))
2/5(sqrt(2-x))^5 - 2(sqrt(2-x))^3
To solve the integration of the function (x+1)sqrt(2-x)dx using the u-substitution method, we need to find an expression for u and its differential du.
Let's make the substitution u = 2-x. To find du, we differentiate both sides of the equation with respect to x:
du/dx = d/dx(2 - x)
The derivative of 2 with respect to x is 0, and the derivative of -x with respect to x is -1, so we have:
du/dx = 0 - 1
du/dx = -1
Now, we can solve for dx in terms of du:
du/dx = -1
dx = -du
Next, rewrite the integral in terms of u and du:
∫(x+1)sqrt(2-x)dx = ∫(x+1)sqrt(u)dx
Since x = 2 - u, we can also express the integral in terms of u and du:
∫(x+1)sqrt(2-x)dx = ∫(2 - u + 1)sqrt(u)(-du)
= ∫(3 - u)sqrt(u)(-du)
Now we can simplify the integrand as (-1)∫(3 - u)sqrt(u)du:
∫(3 - u)sqrt(u)du = -∫(3u - u^2)^(1/2)du
To solve the integral, we can distribute the square root and rewrite the expression:
-∫(3u - u^2)^(1/2)du = -∫((u^2 - 3u)^(1/2))(du)
At this point, the integral becomes easier to solve by using the power rule for integration, which states that:
∫x^n dx = (x^(n+1))/(n+1) + C
Using this rule, where n = 1/2, we can solve the integral:
-∫((u^2 - 3u)^(1/2))(du) = -2/3(u^2 - 3u)^(3/2) + C
Finally, substitute u back in terms of x:
-2/3(u^2 - 3u)^(3/2) + C = -2/3(2 - x)^2 - 3(2 - x)^(3/2) + C
Therefore, the result of the integration is:
∫(x+1)sqrt(2-x)dx = -2/3(2 - x)^2 - 3(2 - x)^(3/2) + C