91Ó°ÊÓ

To evaluate this integral, we can use integration by parts, which is given by the formula: \(\int u dv = uv - \int v du\). We'll choose our functions u and dv as follows: \(u = x^3\): then \(du = 3x^2 dx\), \(dv = e^x dx\): then \(v = e^{x}\). Now, applying integration by parts: \(\int x^{3} e^{x} d x = x^{3} e^{x} - \int e^{x} (3x^{2}) d x\) We need to apply integration by parts one more time: \(u = 3x^2\): then \(du = 6x dx\) \(dv = e^x dx\): then \(v = e^x\) \(\int e^{x} (3x^{2}) d x = 3x^{2} e^{x} - \int e^{x} (6 x) d x\) For the remaining integral, we apply integration by parts for the third time: \(u = 6x\): then \(du = 6 dx\) \(dv = e^x dx\): then \(v = e^x\) \(\int e^{x} (6x) d x = 6x e^{x} - \int e^{x} (6) d x = 6x e^x - 6 \int e^x dx = 6x e^x - 6e^x\) Now, we can substitute this back into our previous results: \(\int x^{3} e^{x} d x = x^{3} e^{x} - \left( 3x^{2} e^{x} - (6x e^x - 6e^x) \right) = x^{3} e^{x} - 3x^{2} e^{x} + 6x e^{x} - 6e^{x}\) So, the solution is \(\int x^{3} e^{x} d x = e^x(x^3 - 3x^2 + 6x - 6) + C\).

To evaluate this integral, we can use integration by parts, in the same way as in integral (i). Choose u and dv as follows: \(u = x^3\): then \(du = 3x^2 dx\) \(dv = \cos x dx\): then \(v = \sin x\) Now, applying integration by parts: \(\int x^{3} \cos x d x = x^{3} \sin x - \int \sin x (3x^{2}) d x\) We need to apply integration by parts again: \(u = 3x^2\): then \(du = 6x dx\) \(dv = \sin x dx\): then \(v = -\cos x\) \(\int \sin x (3x^{2}) d x = -3x^{2} \cos x - \int (-\cos x) (6 x) d x\) For the remaining integral, we apply integration by parts for the third time: \(u = 6x\): then \(du = 6 dx\) \(dv = -\cos x dx\): then \(v = -\sin x\) \(\int (-\cos x) (6x) d x = -6x \sin x - \int (-\sin x) (6) d x = -6x\sin x + 6 \int\sin x dx = -6x \sin x - 6 \cos x\) Now, we can substitute this back into our previous results: \(\int x^{3} \cos x d x = x^{3} \sin x - \left( -3x^{2} \cos x - (-6x \sin x - 6 \cos x) \right) = x^3\sin x + 3x^2\cos x - 6x \sin x + 6\cos x\) So, the solution is \(\int x^{3} \cos x d x = x^3\sin x + 3x^2\cos x - 6x \sin x + 6\cos x + C\).

Notice that we can simplify the expression under the integral sign: \(\frac{x^3}{n^2 x} = \frac{x^3}{x n^2} = \frac{x^2}{n^2}\) Now, the integral becomes: \(\int \frac{x^2}{n^2} dx = \frac{1}{n^2} \int x^2 dx\) Using the power rule for integration, we can write: \(\frac{1}{n^2} \int x^2 dx = \frac{1}{n^2} * \frac{x^3}{3} = \frac{x^3}{3n^2}\) Adding the constant of integration, \(C\), the solution is: \(\int \frac{x^3}{n^2x} dx = \frac{x^3}{3n^2} + C\)