(7) \( \int \frac{d x}{(x+1)^{3}} \) \( \begin{array}{ll}\text { (3) } \int x(3 x-5)^{4} d x & \text { (5) } \int x^{-2 x} d x \\ \text { (7) } \int x^{2} e^{x} d x \\ \text { 1. } \int(x+3 x+3 x-2) d x & \text { (8) } \int 4^{x} e^{2 x} d x \\ =\int & \text { Let } x^{2}\end{array} \)

To tackle the integral \( \int \frac{dx}{(x+1)^{3}} \), you can use substitution or integration rules. A fun trick here is to employ the power rule for integration: rewriting the expression can make it much simpler. If you let \( u = x + 1 \), then \( du = dx \). This transforms your integral into \( \int \frac{du}{u^3} \), leading to a straightforward computation of \( -\frac{1}{2u^2} + C \), and substituting back gives you the final result! Now, did you know that integrals like these are not just academic exercises? They pop up in many real-world scenarios! For instance, when calculating the volume of a liquid in a conical tank, or even in physics when determining the center of mass for irregularly shaped objects. Understanding how to compute these integrals can help you tackle various engineering problems or even design processes, making the abstract math quite tangible and useful!