Question

Evaluate \(\int \frac{x^{3}-6 x^{2}+11 x-6}{\sqrt{x^{2}+4 x+3}} d x\)

Short answer

The integral of the given function is: $$\int \frac{x^{3}-6 x^{2}+11 x-6}{\sqrt{x^{2}+4 x+3}} dx = \frac{1}{4}(x-2)^4 - 2(x-2)^3 + 11(x-2)\sqrt{x^2+4x+3} - 5\mathrm{arcsinh}(x-2) + C.$$

Step-by-step solution

Complete the square

First, let's complete the square for the quadratic expression inside the square root, \(x^2 + 4x + 3\). To do this, we need to find the square of half of the coefficient of \(x\), which is \(2^2 = 4\). Now subtract and add this value inside the expression: $$x^2 + 4x + 3 = (x^2 + 4x + 4) - 4 + 3 = (x+2)^2 - 1.$$

Trigonometric substitution

Now we will use the trigonometric substitution \(x+2 = \sinh(\theta)\) to simplify the square root. By doing this, the square root will become \(\cosh(\theta)\). Also, we need to find the differential \(dx\) in terms of \(d\theta\). Differentiate \(x+2 = \sinh(\theta)\) with respect to \(\theta\): $$\frac{dx}{d\theta} = \frac{d}{d\theta} \sinh(\theta).$$ Thus, \(dx = \cosh(\theta) d\theta\).

Substitute and simplify

Now, make the substitutions \(x + 2 = \sinh(\theta)\) and \(dx = \cosh(\theta) d\theta\) into the given integral: $$\int \frac{x^{3}-6 x^{2}+11 x-6}{\sqrt{x^{2}+4 x+3}} dx = \int \frac{(\sinh(\theta)-2)^{3}-6(\sinh(\theta)-2)^{2}+11(\sinh(\theta)-2)-6}{\cosh(\theta)} \cosh(\theta) d\theta.$$ Simplifying, we get $$\int (\sinh(\theta)-2)^{3}-6(\sinh(\theta)-2)^{2}+11(\sinh(\theta)-2)-6 \, d\theta.$$

Integrate simplified integrand

Next, we'll integrate the simplified integrand with respect to \(\theta\): $$\int (\sinh(\theta)-2)^{3}-6(\sinh(\theta)-2)^{2}+11(\sinh(\theta)-2)-6 \, d\theta = \int (\sinh(\theta)-2)^{3} \, d\theta - 6 \int (\sinh(\theta)-2)^{2} \, d\theta + 11 \int (\sinh(\theta)-2) \, d\theta - 6 \int \, d\theta.$$ Now, integrate each term: $$\int (\sinh(\theta)-2)^{3} \, d\theta = \frac{1}{4} (\sinh(\theta)-2)^{4} + C_1$$ $$\int (\sinh(\theta)-2)^{2} \, d\theta = \frac{1}{3} (\sinh(\theta)-2)^{3} + C_2$$ $$\int (\sinh(\theta)-2) \, d\theta = (\sinh(\theta)-2)\cosh(\theta) + \theta + C_3$$ $$\int \, d\theta = \theta + C_4.$$ Now, we put everything together: $$\frac{1}{4} (\sinh(\theta)-2)^{4} - 6 \left(\frac{1}{3} (\sinh(\theta)-2)^{3}\right) + 11 \left((\sinh(\theta)-2)\cosh(\theta) + \theta\right) - 6\theta + C$$

Reverse substitution and simplification

Finally, substitute back \(x+2 = \sinh(\theta)\): $$\frac{1}{4}(x-2)^4 - 6\left(\frac{1}{3}(x-2)^3\right) + 11\left((x-2)\sqrt{x^2+4x+3} + \mathrm{arcsinh}(x-2)\right) - 6\mathrm{arcsinh}(x-2) + C$$ So, the final answer is: $$\int \frac{x^{3}-6 x^{2}+11 x-6}{\sqrt{x^{2}+4 x+3}} dx = \frac{1}{4}(x-2)^4 - 2(x-2)^3 + 11(x-2)\sqrt{x^2+4x+3} - 5\mathrm{arcsinh}(x-2) + C.$$