$\begin{aligned}\int\frac{2x^{\frac{3}{2}}}{\sqrt{x}}\, dx & =\int\frac{2x^{\frac{3}{2}}}{x^{\frac{1}{2}}}dx\\
& =\int2x^{\frac{3}{2}-\frac{1}{2}}dx\\
& =\int2x\, dx\\
& =x^{2}+c.
\end{aligned}
$

$\int\left(5x^{\frac{2}{3}}+2x^{-\frac{1}{2}}-\sqrt{x}\right)\, dx$$=\int\left(5x^{\frac{2}{3}}+2x^{-\frac{1}{2}}-x^{\frac{1}{2}}\right)\, dx$

$=\frac{5}{\frac{2}{3}+1}x^{\frac{2}{3}+1}+\frac{2}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1}$$-\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}+c$

$=3x^{\frac{5}{3}}+4x^{\frac{1}{2}}-\frac{2}{3}x^{\frac{3}{2}}+c$

$=3\sqrt[3]{x^{5}}+2x^{\frac{1}{2}}\left(2-\frac{1}{3}x\right)+c$

$=3\sqrt[3]{x^{5}}+2\sqrt{x}\left(2-\frac{1}{3}x\right)+c$

$F'(x)=\frac{1}{x^{2}}+1$

$F(x)=\int\left(\frac{1}{x^{2}}+1\right)dx$$=-\frac{1}{x}+x+c$

$F(-1)=0$

$-\frac{1}{(-1)}+(-1)+c=0$$\rightarrow c=0$

Jadi persamaan $F(x)=-\frac{1}{x}+x.$

$\begin{aligned}\int\frac{2x-x^{2}}{x\sqrt{x}}dx & =\int\frac{2x-x^{2}}{x^{\frac{3}{2}}}dx\\
& =\int x^{-\frac{3}{2}}\left(2x-x^{2}\right)dx\\
& =\int\left(2x^{-\frac{1}{2}}-x^{\frac{3}{2}}\right)dx\\
& =4x^{\frac{1}{2}}\left(2-\frac{2}{3}x\right)+c\\
& =2x^{\frac{1}{2}}\left(2-\frac{2}{3}x\right)+c\\
& =2\sqrt{x}\left(\frac{6-2x}{3}\right)+c\\
& =\frac{2}{3}\sqrt{x}\left(6-2x\right)+x
\end{aligned}
$

$\int\left(\frac{x^{5}+3x\sqrt{x}-4}{x^{2}\sqrt{x}}\right)dx$$=\int x^{-\frac{5}{2}}\left(x^{5}+3x\sqrt{x}-4\right)dx$

$=\int\left(x^{-\frac{5}{2}+5}+3x^{\frac{3}{2}+\left(-\frac{5}{2}\right)}-4x^{-\frac{5}{2}}\right)dx$

$=\int\left(x^{\frac{5}{2}}+3x^{-1}-4x^{-\frac{5}{2}}\right)dx$

$=\left(\frac{2}{7}x^{\frac{7}{2}}+3\ln x-\frac{8x}{3}{}^{-\frac{3}{2}}+c\right)$

$=\frac{2}{7}x^{3}\sqrt{x}+3\ln x+\frac{8}{3x\sqrt{x}}+c$

Misalkan $u=6x$$\rightarrow du=6dx$

$dv=\left(3x-1\right)^{-\frac{1}{3}}dx$

$\rightarrow v=\int\left(3x-1\right)^{-\frac{1}{3}}dx$$=\frac{1}{2}\left(3x-1\right)^{\frac{2}{3}}+c$

$\begin{aligned}\int u\, dv & =uv-\int v\mbox{ du}\\
& =6x\cdot\frac{1}{2}\left(3x-1\right)^{\frac{2}{3}}-\int\frac{1}{2}\left(3x-1\right)^{\frac{2}{3}}\cdot6\, dx\\
& =3x\left(3x-1\right)^{\frac{2}{3}}-3\int\left(3x-1\right)^{\frac{2}{3}}dx\\
& =3x\left(3x-1\right)^{\frac{2}{3}}-3\cdot\frac{1}{5}\left(3x-1\right)^{\frac{5}{3}}+c\\
& =3x\left(3x-1\right)^{\frac{2}{3}}-\frac{3}{5}\left(3x-1\right)^{\frac{5}{3}}+c
\end{aligned}
$

Misalkan $u=\left(\frac{1}{2}x+1\right)$$\rightarrow du=\frac{1}{2}dx$

$dv=sin\frac{1}{4}x\, dx$

$v=\int sin\frac{1}{4}x\, dx=-4cos\frac{1}{4}x+c$

$\int u\, dv=uv-\int v\, du$

$\int\left(\frac{1}{2}x+1\right)sin\frac{1}{4}x\mbox{ }dx$$=\left(\frac{1}{2}x+1\right)\cdot-4cos\frac{1}{4}x-\int\frac{1}{2}\cdot\left(-4cos\frac{1}{4}x\right)dx$

$=\left(-2x-4\right)cos\frac{1}{4}x+2\int cos\frac{1}{4}x\, dx$

$=\left(-2x-4\right)cos\frac{1}{4}x+8sin\frac{1}{4}x+c$

$\int_{0}^{2\pi}\left(\frac{1}{2}x+1\right)sin\frac{1}{4}x\mbox{ }dx$$=\left[\left(-2x-4\right)cos\frac{1}{4}x+8sin\frac{1}{4}x\right]_{0}^{2\pi}$

$=\left[\left(-2\cdot2\pi-4\right)cos\frac{2\pi}{4}+8sin\frac{2\pi}{4}\right]$$-\left[\left(-2\cdot0-4\right)cos0+8sin0\right]$

$=\left(8\right)-\left(-4+0\right)$

$=8+4=12.$

Misalkan $u=3x+5$

$du=3\, dx$

$2du=6\, dx$

$\begin{alignedat}{1}\int\frac{6}{3x+5}dx & =\int\frac{2}{u}\, du\\
& =2\ln u+c\\
& =2\ln\left(3x+5\right)+c\\
& =\ln\left(3x+5\right)^{2}+c
\end{alignedat}
$

$\begin{aligned}\int\frac{2x+5}{2x+1}\, dx & =\int\left(1+\frac{4}{2x+1}\right)\, dx\\
& =\int dx+\int\frac{4}{2x+1}\, dx
\end{aligned}
$

Misalkan $u=2x+1$

$du=2$$dx$

$2du=4dx$

$\begin{aligned}\int\frac{2x+5}{2x+1}dx & =\int dx+\int\frac{4}{2x+1}\, dx\\
& =x+2\int\frac{1}{u}\, du\\
& =x+2\ln u+c\\
& =x+2\ln\left|2x+1\right|+c\\
& =x+\ln\left|2x+1\right|^{2}+c
\end{aligned}
$

$\begin{aligned}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{tan^{2}\theta-sin^{2}\theta}} & =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{\frac{sin^{2}\theta}{cos^{2}\theta}-sin^{2}\theta}}\\
& =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{\frac{sin^{2}\theta\left(1-cos^{2}\theta\right)}{cos^{2}\theta}}}\\
& =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{cos\, d\theta}{sin^{2}\theta}\\
& =\left[-\frac{1}{sin\theta}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\\
& =\left(-\frac{1}{sin\frac{\pi}{2}}\right)-\left(-\frac{1}{sin\frac{\pi}{6}}\right)\\
& =-1+2=1
\end{aligned}
$

Perhatikan bahwa $x^{2}+4x+3=(x+3)(x+1)$.

Misalkan $\frac{x-3}{x^{2}+4x+3}=\frac{a}{x+3}+\frac{b}{x+1}$.

Kita dapat;

$\begin{aligned}\frac{x-3}{x^{2}+4x+3} & =\frac{a(x+1)+b(x+3)}{x^{2}+4x+3}\\
& =\frac{(a+b)x+(a+3b)}{x^{2}+4x+3}
\end{aligned}
$

Sehingga, $a+b=1$, dan $a+3b=-3$.

Menyelesaikan persamaan, didapat $a=3,$ $b=-2$.

Jadi

$\int\frac{x-3}{x^{2}+4x+3}\, dx$$=\int\frac{3}{x+3}-\frac{2}{x+1}\, dx$

$=3\ln|x+3|-2\ln|x+1|+c$.

Misalkan $u=1+2e^{x}$

$du=2e^{x}dx$

$\frac{1}{2}du=e^{x}dx$

$\begin{aligned}\int\left(\frac{e^{x}}{1+2e^{x}}\right)dx & =\int\frac{1}{2u}du\\
& =\frac{1}{2}\int\frac{1}{u}du\\
& =\frac{1}{2}\ln\, u+c\\
& =\frac{1}{2}\ln\,\left(1+2e^{x}\right)+c
\end{aligned}
$

Misalkan $u=\ln x$

$du=\frac{1}{x}$$dx$

$\begin{aligned}\int\frac{\ln\, x}{x}dx & =\int u\, du\\
& =\frac{1}{2}u^{2}+c\\
& =\frac{1}{2}\left(\ln\, x\right)^{2}+c
\end{aligned}
$

$\begin{aligned}\frac{1}{x^{2}-4} & =\frac{A}{x+2}+\frac{B}{x-2}\\
& =\frac{A(x-2)+B(x+2)}{x^{2}-4}\\
& =\frac{Ax-2A+Bx+2B}{x^{2}-4}\\
& =\frac{(A+B)x+2B-2A}{x^{2}-4}
\end{aligned}
$

Dari dua persamaan diperoleh :

$A+B=0$…. (1)

$2B-2A=1$…. (2)

Dari pers (1) dan (2) diperoleh :

$A=-\frac{1}{4}$ dan $B=\frac{1}{4}$

$\begin{alignedat}{1}\int\frac{1}{x^{2}-4}d & =\int\frac{-\frac{1}{4}}{x+2}\, dx+\int\frac{\frac{1}{4}}{x-2}\, dx\\
& =-\frac{1}{4}\int\frac{1}{x+2}dx+\frac{1}{4}\int\frac{1}{x-2}\, dx\\
& =-\frac{1}{4}\ln\left(x+2\right)+\frac{1}{4}\ln\left(x-2\right)+c\\
& =\frac{1}{4}\left[\ln\left(\frac{x-2}{x+2}\right)\right]+c
\end{alignedat}
$

$\begin{aligned}\int\frac{2x+1}{x^{2}-3x+2}\, dx & =\int\frac{2x+1}{\left(x-1\right)\left(x-2\right)}\, dx\\
& =\int\frac{A}{x-1}\, dx+\int\frac{B}{x-2}dx
\end{aligned}
$

A dan B dapat dicari dengan menggunakan konsep kesamaan polinom beriku :

$\begin{aligned}\frac{2x+1}{\left(x-1\right)\left(x-2\right)} & =\frac{A}{x-1}+\frac{B}{x-2}\\
& =\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}
\end{aligned}
$

Menurut konsep kesamaan polinom jika ruas kiri sama dengan ruas kanan maka suku-suku sejenisnya adalah
sama. Pada kasusu ini koefisien x dan konstanta pada pembilang.

$2x+1=A(x-2)+B(x-1)$

$2x+1=(x+B)x-2A-B$

$A+B=$2 dan$-2-B=1$

Dengan menggunakan sistem persamaan linear diperoleh A = -3 dan B = 5

$\int\frac{2x+1}{\left(x-1\right)\left(x-2\right)}\, dx=\int\frac{-3}{x-1}dx$$+\int\frac{5}{x-2}dx$

Misalkan $u=x-1\rightarrow du=dx$

$v=x-2\rightarrow dv=dx$

$\int\frac{2x+1}{\left(x-1\right)\left(x-2\right)}\, dx$$=\int\frac{-3}{x-1}dx+\int\frac{5}{x-2}\, dx$

$=\int\frac{-3}{u}du+\int\frac{5}{v}\, dv$

$=-3\ln\, u+5\ln\, v+c$

$=-3\ln\left(x-1\right)+5\ln\left(x-2\right)+c$

$=\ln\frac{\left(x-2\right)^{5}}{\left(x-1\right)^{3}}+c$