Integrate: y/sqrt(2y+1) dx
Thanks!
Actually, the problem was printed in my textbook like that with the equation using the variable y, but with dx rather than dy.
This seems to be a textbook error. I wasn't sure whether that was the case or whether I was doing something wrong.
Why did the mathematician bring a ladder to the calculus party? Because he knew he was going to have to integrate by parts!
Alright, let's tackle this integral step by step.
First, let's make a substitution: let u = 2y + 1. Now, we need to find the derivative of u with respect to y, which is du/dy = 2.
Next, let's rearrange the equation to solve for dy: dy = (1/2)du.
Now we can rewrite the integral in terms of u: ∫(y/√(2y+1))dx = ∫((1/2)du/√u)dx.
Since dx = du/2, we can simplify the integral even further: ∫(1/4√u) du.
Now, let's integrate: ∫(1/4√u) du = (1/4) * ∫u^(-1/2) du = (1/4) * (2u^(1/2)) + C = (u^(1/2))/2 + C.
Finally, substitute u back in terms of y: (2y + 1)^(1/2)/2 + C.
And that's the result! So, the integral of y/√(2y+1) with respect to x is (2y + 1)^(1/2)/2 + C.
Hope that helps, and remember, math parties are always a great place to integrate some fun!
To solve the integral ∫y/√(2y + 1) dx, we can make a substitution to simplify the expression.
Let's substitute u = √(2y + 1). Differentiating both sides of the equation, we get du/dx = (d√(2y + 1))/dx.
To isolate dx, we rearrange the equation as dx = du/(d√(2y + 1))/dx.
Now, let's substitute these new expressions into the integral:
∫y/√(2y + 1) dx = ∫y/u * du/(d√(2y + 1))/dx
Simplifying the expression, we get:
∫y/√(2y + 1) dx = ∫(y/u) * (d√(2y + 1))/dx * dx/du
Notice that (d√(2y + 1))/dx * dx/du is simply du. Therefore, we can rewrite the integral as:
∫y/√(2y + 1) dx = ∫(y/u) du
Now, we can solve this simpler integral. The integral of y/u with respect to u is:
∫(y/u) du = y ∫(1/u) du
Integrating 1/u with respect to u, we get:
∫(1/u) du = ln|u| + C
Now, we substitute u back in terms of y:
ln|u| = ln|√(2y + 1)| = ln(2y + 1)^(1/2) = ln√(2y + 1) = (1/2) ln(2y + 1)
Putting it all together, the solution to the integral is:
∫y/√(2y + 1) dx = y * ln√(2y + 1) + C
where C is the constant of integration.
To integrate the given expression y/sqrt(2y+1) with respect to x, we first need to determine the appropriate substitution to simplify the integral.
Let's make the substitution u = sqrt(2y + 1). To find du/dy, we differentiate both sides of the equation with respect to y:
du/dy = (1/2)*(2y + 1)^(-1/2)*2
Simplifying, we have: du/dy = 1/sqrt(2y + 1).
Now, we need to rewrite the integral in terms of substitution u. To do this, we express dx in terms of du using the equation du/dy:
du/dx = du/dy * dy/dx
Since we want to solve for dx, we rearrange the equation:
dx = du / (du/dx) = du / (du/dy)
Substituting the value we found earlier, dx = du / (1/sqrt(2y + 1)) = sqrt(2y + 1)*du.
Now, we rewrite the integral using the substitution u and dx:
∫ (y/sqrt(2y+1)) dx = ∫ (y/sqrt(2y+1)) * sqrt(2y + 1) du
Simplifying further, the square root term cancels out:
∫ y du
Finally, we can integrate the expression:
∫ y du = (1/2)u^2 + C.
Replacing the value of u back into the equation, we have the final answer:
∫ (y/sqrt(2y+1)) dx = (1/2)*(sqrt(2y + 1))^2 + C = (1/2)*(2y + 1) + C = y + 1/2 + C.
You must mean y/sqrt(2y+1) dy
Use the substitution 2y + 1 = u
dy = du/2
y = (u-1)/2
The integral becomes
(Integral of) [(1/2)(u-1)/sqrt u]du/2
= (Integral of) (1/4) [sqrt u - 1/sqrt u] du
You finish it.