question 4\n4.1 determine the following integrals:\n4.1.1 $int\frac{3x^{2}+sin x}{x^{3}-cos…

Explanation:

Step1: Solve 4.1.1

Let $u = x^{3}-\cos x$, then $du=(3x^{2}+\sin x)dx$. So $\int\frac{3x^{2}+\sin x}{x^{3}-\cos x}dx=\int\frac{du}{u}=\ln|u|+C=\ln|x^{3}-\cos x|+C$.

Step2: Solve 4.1.2

First, complete the square for the denominator: $x^{2}-2x=(x - 1)^{2}-1$. Let $u=x - 1$, then $x=u + 1$ and $dx=du$. The integral becomes $\int\frac{u}{ \sqrt{u^{2}-1}}du+\int\frac{- 1}{\sqrt{u^{2}-1}}du$. For $\int\frac{u}{\sqrt{u^{2}-1}}du$, let $t = u^{2}-1$, $dt = 2udu$, so $\int\frac{u}{\sqrt{u^{2}-1}}du=\sqrt{u^{2}-1}+C_1=\sqrt{x^{2}-2x}+C_1$. For $\int\frac{-1}{\sqrt{u^{2}-1}}du=-\ln|u+\sqrt{u^{2}-1}|+C_2=-\ln|(x - 1)+\sqrt{x^{2}-2x}|+C_2$. The result is $\sqrt{x^{2}-2x}-\ln|(x - 1)+\sqrt{x^{2}-2x}|+C$.

Step3: Solve 4.1.3

Let $t=2x^{3}-4$, then $dt = 6x^{2}dx$, $x^{2}dx=\frac{1}{6}dt$. When $x = 1$, $t=2\times1^{3}-4=-2$; when $x = 2$, $t=2\times2^{3}-4 = 12$. The integral $\int_{1}^{2}x^{2}(2x^{3}-4)^{3}dx=\frac{1}{6}\int_{-2}^{12}t^{3}dt=\frac{1}{6}\times\frac{t^{4}}{4}\big|_{-2}^{12}=\frac{1}{24}(12^{4}-(-2)^{4})=\frac{1}{24}(20736 - 16)=\frac{20720}{24}=\frac{2590}{3}$.

Step4: Solve 4.1.4

$\int(\frac{5}{x}+e^{5x})dx=5\int\frac{1}{x}dx+\int e^{5x}dx$. Since $\int\frac{1}{x}dx=\ln|x|+C_1$ and $\int e^{5x}dx=\frac{1}{5}e^{5x}+C_2$, the result is $5\ln|x|+\frac{1}{5}e^{5x}+C$.

Answer:

4.1.1: $\ln|x^{3}-\cos x|+C$ 4.1.2: $\sqrt{x^{2}-2x}-\ln|(x - 1)+\sqrt{x^{2}-2x}|+C$ 4.1.3: $\frac{2590}{3}$ 4.1.4: $5\ln|x|+\frac{1}{5}e^{5x}+C$