Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant of integration.)
1/x{(x-1)^1/2} dx
Why did the integral go to the school dance? Because it wanted to find a rational function to slow dance with!
Let's make a substitution, u = x - 1, to express the integrand as a rational function.
Now, du = dx, and when we substitute these values into the integral, we get:
∫ (1/x) * (x - 1)^(1/2) dx = ∫ (1/(u+1)) * u^(1/2) du
Now, let's simplify this expression:
∫ (1/(u+1)) * u^(1/2) du = ∫ (u^(1/2))/(u + 1) du
Using the power rule of integration, we can rewrite this as:
∫ (u^(1/2))/(u + 1) du = 2√u - 2ln|√u + 1| + C
Finally, substituting back u = x - 1, we get:
2√(x - 1) - 2ln|√(x - 1) + 1| + C
And that's the result of evaluating the integral!
To express the integrand as a rational function, we can make the substitution:
Let u = √(x-1)
To find du, we differentiate both sides of this equation with respect to x:
du/dx = 1/(2√(x-1))
Solving for dx, we find:
dx = 2√(x-1) du
Now we can rewrite the integrand in terms of u:
1/x√(x-1) dx = 1/(x * u) * 2√(x-1) du = 2 du/u
The integral becomes:
∫ 2 du/u
Integrating this expression, we get:
2 ∫ du/u = 2 ln|u| + C
Recall that u = √(x-1):
2 ln|√(x-1)| + C
Finally, we simplify the natural logarithm expression:
= ln|(x-1)| + C
To express the integrand as a rational function, we can make a substitution.
Let's set u = sqrt(x-1), which means that x = u^2 + 1.
Now, let's find the derivative of x with respect to u in order to substitute the variables correctly.
dx = 2u du
Substituting these values into the original integral, we have:
∫ 1/x * sqrt(x-1) dx
= ∫ 1/(u^2 + 1) * u * 2u du
= 2∫ u/(u^2 + 1) du
To evaluate this integral, we can use the substitution method once again. Let v = u^2 + 1, which means that du = (1/2)dv.
Substituting these values, we get:
2∫ (1/2(v-1)) dv
= ∫ (1/2v - 1/2) dv
= (1/2) ∫ (1/v) dv - (1/2) ∫ (1) dv
= (1/2) ln|v| - (1/2)v + C
Remember, v = u^2 + 1, so let's substitute back:
= (1/2) ln|u^2 + 1| - (1/2)(u^2 + 1) + C
= (1/2) ln|sqrt(x-1)^2 + 1| - (1/2)(sqrt(x-1)^2 + 1) + C
= (1/2) ln|x-1 + 1| - (1/2)(x-1 + 1) + C
= (1/2) ln|x| - (1/2)(x-1) + C
= (1/2) ln|x| - (1/2)x + (1/2) + C
Thus, the integral of 1/x * sqrt(x-1) dx is (1/2) ln|x| - (1/2)x + (1/2) + C.
let
u^2 = x-1
x = 1+u^2
2u du = dx
Now your integrand is
1/(1+u^2) * u * 2u du
= 2u^2/(1+u^2) du
= 2(1 - 1/(1+u^2)) du
That integrates to
2(u - arctan(u))+C
= 2(√(x-1) - arctan√(x-1))+C