∫x√x2+1

Sure, I can help with that. The integral you're asking about is a bit tricky, but we can solve it using a substitution method. Here are the steps:

1. Let's set u = x^2 + 1. Then, du = 2x dx.

2. But we have x dx in our integral, not 2x dx. So, we'll divide both sides of du = 2x dx by 2 to get (1/2)du = x dx.

3. Now we can substitute u and du into our integral. The integral becomes ∫√u * (1/2) du.

4. We can pull the 1/2 out of the integral to get (1/2) ∫√u du.

5. The integral of √u with respect to u is (2/3)u^(3/2).

6. So, our integral becomes (1/2) * (2/3)u^(3/2) = (1/3)u^(3/2).

7. Finally, we substitute x^2 + 1 back in for u to get our final answer: (1/3)(x^2 + 1)^(3/2) + C, where C is the constant of integration.

To solve the integral ∫x√(x^2+1) dx, we can use the method of substitution. Here are the steps:

Step 1: Choose a substitution. In this case, we can let u = x^2 + 1.

Step 2: Differentiate the substitution to find du. The derivative of x^2 + 1 is 2x, so du = 2x dx.

Step 3: Solve the equation du = 2x dx for dx. We get dx = du / (2x).

Step 4: Substitute u and dx into the integral. We get ∫x√u du / (2x).

Step 5: Simplify the integral. The x's cancel out, and we get (1/2)∫√u du.

Step 6: Now we can integrate. The integral of √u with respect to u is (2/3)u^(3/2), so we get (1/2)*(2/3)u^(3/2) = (1/3)u^(3/2).

Step 7: Substitute x^2 + 1 back in for u. We get (1/3)(x^2 + 1)^(3/2).

Step 8: Don't forget the constant of integration! The final answer is (1/3)(x^2 + 1)^(3/2) + C.

This is an integral calculus problem. Here's how you can solve it:

Step 1: Recognize the integral as a standard form The integral ∫x√(x^2+1) dx can be recognized as a standard form of integration by substitution. The standard form is ∫f'(x)√(f(x)^2+a^2) dx, where f(x) = x, f'(x) = 1, and a = 1.

Step 2: Apply the formula The formula for this standard form is (1/2a) * f(x) * √(f(x)^2+a^2) + (a^2/2) * ln|f(x) + √(f(x)^2+a^2)| + C.

Step 3: Substitute the values Substitute f(x) = x, f'(x) = 1, and a = 1 into the formula. This gives (1/2) * x * √(x^2+1) + (1/2) * ln|x + √(x^2+1)| + C.

So, the integral ∫x√(x^2+1) dx = (1/2) * x * √(x^2+1) + (1/2) * ln|x + √(x^2+1)| + C.