Question

Evaluate the following integrals. $$\int \frac{x^{2}+2 x+4}{\sqrt{x^{2}-4 x}} d x, x>4$$

Short answer

To summarize, the integral of the given function is: $$\int \frac{x^2 + 2x + 4}{\sqrt{x^2-4x}} dx = \frac{3}{2}(x-2)^2 + 3\sinh{(2\cosh^{-1}(\frac{x-2}{2}))} - 6\cosh^{-1}(\frac{x-2}{2}) + C$$

Step-by-step solution

Simplify the given function and identify a substitution

We are given the integral to evaluate as: $$\int \frac {x^2 + 2x + 4}{\sqrt{x^2 - 4x}} dx$$ We first try to simplify the denominator by completing the square. The expression \(x^2-4x\) can be written as \((x-2)^2 - 4\). Now, we have the expression as: $$\int \frac {x^2 + 2x + 4}{\sqrt{(x-2)^2 - 4}} dx$$ We can substitute \(u = x - 2\), then the derivative of \(u\) with respect \(x\) is \(du = dx\).

Implement the substitution

Replacing \(x\) with \(u + 2\) in the function, the integral becomes: $$\int \frac {(u+2)^2 + 2(u+2) + 4}{\sqrt{u^2 - 4}} du$$ Expanding and simplifying the numerator yields $$\int \frac {u^2 + 4u + 4 + 2u + 4}{\sqrt{u^2 - 4}} du$$ Which simplifies to $$\int \frac {u^2 + 6u + 12}{\sqrt{u^2 - 4}} du$$

Break down the function using partial fraction decomposition

Now we can perform partial fraction decomposition for the integrand. Let's find the coefficients \(A\) and \(B\) such that: $$\frac {u^2 + 6u + 12}{\sqrt{u^2 - 4}} = A\left(u + \sqrt{u^2-4}\right) + B\left(u - \sqrt{u^2-4}\right)$$ We'll focus on the numerator to find \(A\) and \(B\). Equating coefficients we get: $$(A+B)u + (A-B)\sqrt{u^2 - 4} = u^2 + 6u + 12$$ Comparing like terms, we get: $$A + B = 6$$ $$A-B= 0$$ Solving this system of equations, we find that \(A=B=3\). Our decomposed integral becomes: $$\int 3\left(u + \sqrt{u^2-4}\right) du + \int 3\left(u - \sqrt{u^2-4}\right) du$$

Evaluate the indefinite integrals

Now, we can evaluate the indefinite integrals separately: $$I_1 = \int 3\left(u + \sqrt{u^2-4}\right) du = 3\int u du + 3\int \sqrt{u^2 - 4} du$$ First integral: $$3\int u du = 3\frac{1}{2}u^2 = \frac{3}{2}u^2$$ Second integral: To solve this integral, we'll let \(u = 2\cosh{(t)}\), with \(du = 2\sinh{(t)}dt\). The integral becomes: $$3\int 2\sinh{(t)}\sqrt{4\cosh^2{(t)}-4} dt = 6\int \sinh{(t)}\sqrt{4(\cosh^2{(t)}-1)}dt$$ Using the identity \(\cosh^2(t) - 1 = \sinh^2(t)\), we get: $$6\int \sinh{(t)}\cdot 2\sinh{(t)} dt = 12\int \sinh^2{(t)} dt$$ Using the identity \(\sinh^2(t) = \frac{1}{2}(\cosh(2t) - 1)\), the integral becomes: $$12\int \frac{1}{2}(\cosh{(2t)}-1) dt = 6\int (\cosh{(2t)} - 1) dt = 6\int\cosh{(2t)} dt - 6\int dt$$ Now, we evaluate the integrals: $$6\int\cosh{(2t)} dt - 6\int dt = 3\frac{1}{2}\sinh{(2t)} - 6t + C = 3\sinh{(2t)} - 6t + C$$ Using the inverse of the function \(u=2\cosh{(t)}\), namely \(t = \cosh^{-1}(\frac{u}{2})\), we finally find that: $$3\sinh{(2t)} - 6t + C = 3\sinh{(2\cosh^{-1}(\frac{u}{2}))} - 6\cosh^{-1}(\frac{u}{2}) + C$$ So we get $$I_1 = \frac{3}{2}u^2 + 3\sinh{(2\cosh^{-1}(\frac{u}{2}))} - 6\cosh^{-1}(\frac{u}{2}) + C_1$$ The integral \(I_2\) can be found using the same steps. We can then sum up the results to find the general solution.

Add the evaluated integrals and substitute the original variable

Now, sum up the integrals \(I_1\) and \(I_2\) and then substitute back the original variable, \(x = u + 2\): $$\int \frac {x^2 + 6x + 12}{\sqrt{x^2 - 4x}} dx = I_1+I_2$$ Replace \(u\) by \(x-2\) in the sum and simplify to obtain the final result: $$\int \frac{x^2 + 2x + 4}{\sqrt{x^2-4x}} dx = \frac{3}{2}(x-2)^2 + 3\sinh{(2\cosh^{-1}(\frac{x-2}{2}))} - 6\cosh^{-1}(\frac{x-2}{2}) + C$$