Question

Q. Intigrate
ʃ[x4/(x-1)(x+2)1/2]dx

Answer 1

Answer on Question #53148 – Math – Integral Calculus

Integrate $\int \frac{x^4}{(x - 1)\sqrt{x + 2}} dx$

Solution

$$\begin{aligned} \int \frac{x^4}{(x - 1)\sqrt{x + 2}} dx &= \left| \begin{array}{cc} \sqrt{x + 2} = t & x^4 = (t^2 - 2)^4 \\ dx = 2t\,dt & x - 1 = t^2 - 3 \end{array} \right| = 2 \int \frac{(t^2 - 2)^4}{t^2 - 3} dt = \\ &= 2 \int \frac{t^8 - 8t^6 + 24t^4 - 32t^2 + 16}{t^2 - 3} dt = \\ &= 2 \int \frac{t^6(t^2 - 3) + 5t^4(t^2 - 3) + 9t^2(t^2 - 3) - 5(t^2 - 3) + 1}{t^2 - 3} dt = \end{aligned}$$ $$\begin{aligned} &= 2 \int \left(t^6 - 5t^4 + 9t^2 - 5 + \frac{1}{t^2 - 3}\right) dt = 2 \left(\frac{t^7}{7} - t^5 + 3t^3 - 5t + \frac{1}{2\sqrt{3}} \log \left| \frac{t - \sqrt{3}}{t + \sqrt{3}} \right|\right) + C = \\ &= 2t \left(\frac{t^6}{7} - t^4 + 3t^2 - 5\right) + \frac{1}{\sqrt{3}} \log \left| \frac{t - \sqrt{3}}{t + \sqrt{3}} \right| + C = \\ &= 2\sqrt{x + 2} \left(\frac{(x + 2)^3}{7} - (x + 2)^2 + 3(x + 2) - 5\right) + \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} \right| + C = \\ \end{aligned}$$ $$= \frac{2}{7} \sqrt{x + 2} (x^3 - x^2 + 5x - 13) + \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} \right| + C,$$

where $C$ is an arbitrary real constant.

**Answer:** $\frac{2}{7} \sqrt{x + 2} (x^3 - x^2 + 5x - 13) + \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} \right| + C.$

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