Question

solve the following:
① $int\frac{3x^{2}}{5x^{3}-1}dx$
② $int\frac{2 - 3sin x}{cos^{2}x}dx$
③ $int\frac{ln^{2}3x}{x}dx$
④ $int\frac{\tan(ln x)}{x}dx$
⑤ $int\frac{2+ln^{3}x}{x(1 - ln x)}dx$
⑥ $int\frac{sin3t}{cos3t - 1}dt$
⑦ $int\frac{1}{sqrt{x}(4+sqrt{x})}dx$
⑧ $int_{1}^{5}\frac{4z^{3}-1}{2z - 1}dz$

Explanation:

Step1: For $\int\frac{3x^{2}}{5x^{2}-1}dx$

First, rewrite as $\int\frac{\frac{3}{5}(5x^{2}-1)+\frac{3}{5}}{5x^{2}-1}dx=\int(\frac{3}{5}+\frac{3}{5(5x^{2}-1)})dx$. Then $\int\frac{3}{5}dx+\frac{3}{5}\int\frac{1}{5x^{2}-1}dx$. The first integral is $\frac{3}{5}x$. For the second, use partial - fraction decomposition on $\frac{1}{5x^{2}-1}=\frac{1}{(\sqrt{5}x - 1)(\sqrt{5}x+1)}=\frac{1}{2\sqrt{5}}(\frac{1}{\sqrt{5}x - 1}-\frac{1}{\sqrt{5}x + 1})$. Integrating gives $\frac{3}{5}x+\frac{3}{10\sqrt{5}}\ln|\frac{\sqrt{5}x - 1}{\sqrt{5}x + 1}|+C$.

Step2: For $\int\frac{2 - 3\sin x}{\cos^{2}x}dx$

Split the integral: $\int\frac{2}{\cos^{2}x}dx-3\int\frac{\sin x}{\cos^{2}x}dx$. We know that $\int\frac{1}{\cos^{2}x}dx=\tan x$ and for $\int\frac{\sin x}{\cos^{2}x}dx$, let $u = \cos x$, $du=-\sin xdx$, then $\int\frac{\sin x}{\cos^{2}x}dx=-\int u^{-2}du=\frac{1}{u}+C=\frac{1}{\cos x}+C$. So the result is $2\tan x + \frac{3}{\cos x}+C$.

Step3: For $\int\frac{\ln^{2}3x}{x}dx$

Let $u=\ln(3x)$, $du=\frac{1}{x}dx$. Then the integral becomes $\int u^{2}du=\frac{1}{3}u^{3}+C=\frac{1}{3}\ln^{3}(3x)+C$.

Step4: For $\int\frac{\tan(\ln x)}{x}dx$

Let $u = \ln x$, $du=\frac{1}{x}dx$. Then the integral is $\int\tan udu=-\ln|\cos u|+C=-\ln|\cos(\ln x)|+C$.

Step5: For $\int\frac{2+\ln^{3}x}{x(1 - \ln x)}dx$

Let $t=\ln x$, $dt=\frac{1}{x}dx$. The integral becomes $\int\frac{2 + t^{3}}{1 - t}dt$. First, perform polynomial long - division: $\frac{2 + t^{3}}{1 - t}=-t^{2}-t - 1+\frac{1}{1 - t}$. Integrating term - by - term gives $-\frac{1}{3}t^{3}-\frac{1}{2}t^{2}-t-\ln|1 - t|+C=-\frac{1}{3}\ln^{3}x-\frac{1}{2}\ln^{2}x-\ln x-\ln|1 - \ln x|+C$.

Step6: For $\int\frac{\sin3t}{\cos3t - 1}dt$

Let $u=\cos3t - 1$, $du=-3\sin3tdt$. Then the integral is $-\frac{1}{3}\int\frac{du}{u}=-\frac{1}{3}\ln|\cos3t - 1|+C$.

Step7: For $\int\frac{1}{\sqrt{x}(4+\sqrt{x})}dx$

Let $u = 4+\sqrt{x}$, $du=\frac{1}{2\sqrt{x}}dx$. Then the integral is $2\int\frac{du}{u}=2\ln|4+\sqrt{x}|+C$.

Step8: For $\int_{1}^{5}\frac{4z^{3}-1}{2z - 1}dz$

First, perform polynomial long - division on $\frac{4z^{3}-1}{2z - 1}=2z^{2}+z+\frac{1}{2}+\frac{- \frac{1}{2}}{2z - 1}$. Then integrate term - by - term: $\int(2z^{2}+z+\frac{1}{2}-\frac{1}{2(2z - 1)})dz=\frac{2}{3}z^{3}+\frac{1}{2}z^{2}+\frac{1}{2}z-\frac{1}{4}\ln|2z - 1|$. Evaluate the definite integral: $(\frac{2}{3}(5)^{3}+\frac{1}{2}(5)^{2}+\frac{1}{2}(5)-\frac{1}{4}\ln|9|)-(\frac{2}{3}(1)^{3}+\frac{1}{2}(1)^{2}+\frac{1}{2}(1)-\frac{1}{4}\ln|1|)=\frac{250}{3}+\frac{25}{2}+\frac{5}{2}-\frac{1}{4}\ln9-\frac{2}{3}- \frac{1}{2}-\frac{1}{2}=\frac{248}{3}+14-\frac{1}{4}\ln9=\frac{248 + 42}{3}-\frac{1}{4}\ln9=\frac{290}{3}-\frac{1}{4}\ln9$.

Answer:

The solutions for the integrals are:

1. $\frac{3}{5}x+\frac{3}{10\sqrt{5}}\ln|\frac{\sqrt{5}x - 1}{\sqrt{5}x + 1}|+C$
2. $2\tan x+\frac{3}{\cos x}+C$
3. $\frac{1}{3}\ln^{3}(3x)+C$
4. $-\ln|\cos(\ln x)|+C$
5. $-\frac{1}{3}\ln^{3}x-\frac{1}{2}\ln^{2}x-\ln x-\ln|1 - \ln x|+C$
6. $-\frac{1}{3}\ln|\cos3t - 1|+C$
7. $2\ln|4+\sqrt{x}|+C$
8. $\frac{290}{3}-\frac{1}{4}\ln9$