# How do you integrate #int (x+1)sqrt(2-x)dx#?

Factoring and making it look nice:

By signing up, you agree to our Terms of Service and Privacy Policy

To integrate ( \int (x+1)\sqrt{2-x} , dx ), you can use the substitution method. Let ( u = 2-x ). Then, ( du = -dx ).

So, ( dx = -du ), and ( x = 2 - u ).

Substituting these into the integral:

( \int (2 - u + 1)\sqrt{u} \cdot (-du) )

( = -\int (3 - u)\sqrt{u} , du )

Expand the expression:

( = -\int (3\sqrt{u} - u\sqrt{u}) , du )

Now integrate each term separately:

( = -\left(3\int \sqrt{u} , du - \int u\sqrt{u} , du\right) )

( = -\left(3 \cdot \frac{2}{3}u^{3/2} - \frac{2}{5}u^{5/2}\right) + C )

( = -2u^{3/2} + \frac{2}{5}u^{5/2} + C )

Now, revert ( u ) back to ( x ):

( = -2(2-x)^{3/2} + \frac{2}{5}(2-x)^{5/2} + C )

This is the integral of ( (x+1)\sqrt{2-x} ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7