Integrate 1/√x(1+x) dx
Let's start by making a substitution:
u = 1 + x (so du/dx = 1 and dx = du)
Then the integral becomes:
∫ 1/√x(1+x) dx = ∫ 1/√u du
Using the power rule of integration for u^(n+1)/(n+1) with n=-1/2:
∫ 1/√u du = 2√u + C
Substituting back to the original variable x:
2√(1+x) + C = 2√(x+1) + C
Therefore, the indefinite integral of 1/√x(1+x) dx is 2√(x+1) + C.
To integrate the expression 1/√x(1+x) dx, we can use a technique called substitution.
Step 1: Let's start by making a substitution. Let u = √x. This means that u^2 = x, and we can express dx in terms of du.
Differentiating both sides of the equation u^2 = x with respect to x, we get:
2u * du = dx
Now we can substitute these expressions in the original integral.
Step 2: Rewrite the integral using the substitution:
∫ 1/√x(1+x) dx = ∫ 1/u * (1 + u^2) * 2u * du
Simplifying further, we get:
∫ 2(1 + u^2) du/u
Step 3: Distribute the 2 across the terms inside the integral:
∫ 2 du/u + ∫ 2u^2 du/u
Simplifying further, we get:
∫ 2 du/u + ∫ 2u du
Step 4: Integrate each term separately.
The first term, ∫ 2 du/u, simplifies to 2ln|u| + C, where C is the constant of integration.
For the second term, ∫ 2u du, we can use the power rule for integration. Integrating u with respect to u gives u^2/2.
Therefore, the second term simplifies to u^2 + C.
Step 5: Substitute back for u in terms of x.
Since we let u = √x, u^2 = x.
So the integral becomes:
2ln|u| + u^2 + C
Substituting u = √x, we have:
2ln|√x| + (√x)^2 + C
Simplifying further, we have:
2ln|√x| + x + C
Therefore, the integral of 1/√x(1+x) dx is equal to:
2ln|√x| + x + C, where C is the constant of integration.