Question

Integrate with respect to x :
∫2−1x^2/(x^3+4)^2dx
2 superscrit while -1 subscript

12
1/2
6
5/12

Answer 1

Answer on Question #46131 – Math – Integral Calculus

Integrate with respect to $x$:

$$\int_{-1}^{2} \frac{x^2}{(x^3 + 4)^2} dx$$

Solution:

We’ll make a substitution $t = x^3 + 4$:

$$\begin{array}{l} \int_{-1}^{2} \frac{x^2}{(x^3 + 4)^2} dx = \left| \begin{array}{c} t = x^3 + 4 \\ dt = 3x^2 dx \\ t(x = -1) = 3 \\ t(x = 2) = 12 \end{array} \right| = \frac{1}{3} \int_{-1}^{2} \frac{3x^2 dx}{(x^3 + 4)^2} = \frac{1}{3} \int_{3}^{12} \frac{dt}{t^2} = \frac{1}{3} \left(-\frac{1}{t}\right) \Bigg|_{t=3}^{t=12} \\ = -\frac{1}{3} \left(\frac{1}{12} - \frac{1}{3}\right) = -\frac{1}{3} \left(-\frac{3}{12}\right) = \frac{1}{12} \end{array}$$

Answer: 1/12

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