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

$dv=\sqrt{x+2}dx$

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

$\begin{aligned}\int udv & =uv-\int vdu\\
\int x\sqrt{x+2} & =x\cdot\frac{2}{3}\left(x+2\right)^{\frac{3}{2}}-\int\frac{2}{3}\left(x+2\right)^{\frac{3}{2}}dx\\
& =x\cdot\frac{2}{3}\left(x+2\right)^{\frac{3}{2}}-\frac{2}{3}\cdot\frac{2}{5}\left(x+2\right)^{\frac{5}{2}}+C\\
& =\frac{2}{3}\left(x+2\right)^{\frac{3}{2}}\left\{ x-\frac{2}{5}\left(x+2\right)\right\} +C\\
& =\frac{2}{3}\left(x+2\right)^{\frac{3}{2}}\left\{ \frac{5x-2(x+2)}{5}\right\} +C\\
& =\frac{2}{3}\left(x+2\right)^{\frac{3}{2}}\left(\frac{3x-2}{5}\right)+C\\
& =\frac{2}{15}\left(3x-2\right)\left(x+2\right)\sqrt{x+2}+C
\end{aligned}
$

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\, 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=x$$\rightarrow du=dx$

$dv=\left(x+3\right)^{4}dx$

$\begin{aligned}\rightarrow v & =\int\left(x+3\right)^{4}dx\\
& =\frac{1}{5}(x+3)^{5}+c
\end{aligned}
$

$\begin{aligned}\int udv & =uv-\int vdu\\
\int x\left(x+3\right)^{4}\, dx & =x\cdot\frac{1}{5}(x+3)^{5}-\int\frac{1}{5}(x+3)^{5}\, dx\\
& =\frac{x}{5}(x+3)^{5}-\frac{1}{30}\left(x+3\right)^{6}+c\\
& =\frac{1}{5}\left(x+3\right)^{5}\left\{ x-\frac{1}{6}\left(x+3\right)\right\} +c\\
& =\frac{1}{5}\left(x+3\right)^{5}\left(\frac{6x-x-3}{6}\right)+c\\
& =\frac{1}{30}(x+3)^{5}(5x-3)+c
\end{aligned}
$

Misalkan $u=x\rightarrow du=dx$

$dv=\frac{1}{\sqrt{x+4}}dx$

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

$\begin{aligned}\int u\, dv & =uv-\int v\, du\\
\int\frac{x}{\sqrt{x+4}}\mbox{dx} & =x\cdot2\left(x+4\right)^{\frac{1}{2}}-\int2\left(x+4\right)^{\frac{1}{2}}+c\\
& =2x\left(x+4\right)^{\frac{1}{2}}-\frac{4}{3}\left(x+4\right)^{\frac{3}{2}}+c\\
& =2\left(x+4\right)^{\frac{1}{2}}\left[x-\frac{2}{3}\left(x+4\right)\right]+c\\
& =2\sqrt{x+4}\left(\frac{3x-2x-8}{3}\right)+c\\
& =2\sqrt{x+4}\left(\frac{x-8}{3}\right)+c\\
& =\frac{2}{3}\left(x-8\right)\sqrt{x+4}+c
\end{aligned}
$

Misalkan $u=10x$ $\rightarrow du=10\, dx$

$dv=\frac{1}{\sqrt[4]{2x-1}}dx$

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

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

$\int\frac{10x}{\sqrt[4]{2x-1}}\, dx$$=10x\cdot\frac{2}{3}\left(2x-1\right)^{\frac{3}{4}}-\int10\cdot\frac{2}{3}\left(2x-1\right)^{\frac{3}{4}}dx$

$=\frac{20x}{3}\left(2x-1\right)^{\frac{3}{4}}-\frac{20}{3}\int\left(2x-1\right)^{\frac{3}{4}}dx$

$=\frac{20x}{3}\left(2x-1\right)^{\frac{3}{4}}-\frac{20}{3}\cdot\frac{2}{7}\left(2x-1\right)^{\frac{7}{4}}+c$

$=\frac{20}{3}\left(2x-1\right)^{\frac{3}{4}}\left\{ x-\frac{2}{7}\left(2x-1\right)\right\} +c$

$=\frac{20}{3}\left(2x-1\right)^{\frac{3}{4}}\left(\frac{7x-4x+2}{7}\right)+c$

$=\frac{20x}{3}\left(2x-1\right)^{\frac{3}{4}}\left(\frac{3x+2}{7}\right)+c$

$=\frac{20x}{21}(3x+2)\left(2x-1\right)^{\frac{3}{4}}+c.$

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

$dv=cosx\, dx$

$\begin{aligned}v & =\int cosx\, dx\\
& =sinx+c
\end{aligned}
$

$\begin{aligned}\int udv & =uv-\int vdu\\
\int x\cdot cosx\, dx & =x\cdot sinx-\int sinx\, dx\\
& =x\cdot sinx+cosx+c
\end{aligned}
$

$\int_{0}^{\frac{\pi}{2}}x\cdot cosx\mbox{ }dx$$=\left[x\cdot sinx+cosx\right]_{0}^{\frac{\pi}{2}}$

$=\left(\frac{\pi}{2}\cdot sin\frac{\pi}{2}+cos\frac{\pi}{2}\right)-\left(0\cdot sin0+cos0\right)$

$=\left(\frac{\pi}{2}\right)-1.$

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

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

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

$\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\left(-4cos\frac{1}{4}x\right)-\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=2x^{2}$$\rightarrow du=4x\, dx$

$dv=sin2x\, dx$

$v=\int sin2x\, dx$$=-\frac{1}{2}cos2x+c$

$\int udv=uv-\int vdu$

$=2x^{2}\cdot\left(-\frac{1}{2}cos2x\right)$$-\int4x\cdot\left(-\frac{1}{2}cos2x\right)dx+c$

$=-x^{2}\cdot cos2x+2\int x\cdot cos2x\, dx+c$

$\int x\cdot cos2x\, dx$ cari dengan integral parsial juga :

$\begin{aligned}\int x\cdot cos2x\, dx & =x\left(\frac{1}{2}sin2x\right)-\int\frac{1}{2}sin2x\, dx\\
& =\frac{x}{2}sin2x+\frac{1}{4}cos2x+c
\end{aligned}
$

$\int2x^{2}\cdot sin2x\, dx$$=-x^{2}\cdot cos2x+2\int x\cdot cos2x\, dx+c$

$=-x^{2}\cdot cos2x+2\left(\frac{x}{2}sin2x+\frac{1}{4}cos2x\right)+c$

$=-x^{2}\cdot cos2x+x\cdot sin2x+\frac{1}{2}cos2x+c$.

Misalkan : $u=(x^{2}+1)\rightarrow du=2x\, dx$

$dv=cosx$$dx$

$\rightarrow v=\int cosxdx=sinx+c$

$\int udv=uv-\int vdu$

$\int\left(x^{2}+1\right)cosx\, dx$$=\left(x^{2}+1\right)sinx-\int sinx\,2x\, dx $

untuk $\int2x\cdot sinx\, dx$ harus menggunakan integral parsial lagi :

$\begin{aligned}\int2x\cdot sinx\, dx & =2x\cdot(-cosx)-\int2\cdot(-cosx)dx\\
& =-2xcosx+2sinx+c
\end{aligned}
$

$\int\left(x^{2}+1\right)cosxdx$$=\left(x^{2}+1\right)sinx-\int sinx\,2x\, dx$

$=\left(x^{2}+1\right)sinx-\left(-2xcosx+2sinx\right)+c$

$=\left(x^{2}+1\right)sinx+2x\cdot cosx-2sinx+c$

$=\left(x^{2}+1-2\right)sinx+2x\cdot cosx+c$

$=\left(x^{2}-1\right)sinx+2x\cdot cosx+c.$

Misalkan $u=3x$$\rightarrow du=3dx$

$dv=sec^{2}6x$$dx$

$v=\int sec^{2}6x\, dx=\frac{1}{6}tan\,6x+c$

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

$\int3x\cdot sec^{2}6x\, dx$$=3x\cdot\frac{1}{6}tan6x-\int3\cdot\frac{1}{6}tan6x\, dx$

$=\frac{x}{2}tan6x-\frac{1}{2}\int tan\,6x\, dx$

$=\frac{x}{2}tan6x-\frac{1}{12}\ln\left|sec\,6x\right|+c$

$=\frac{x}{2}tan6x-\frac{1}{12}\ln\left|sec\,6x\right|+c.$

Ambil,

$u=x$

$dv=\sin(x)dx$

$du=dx$

$v=-\cos(x)$

Maka

$\begin{aligned}\int x\sin(x)dx & =uv-\int v\, du\\
& =-x\cos(x)+\int\cos(x)dx\\
& =-x\cos(x)+\sin(x)+c.
\end{aligned}
$

Ambil,

$u=x$

$dv=\sin(x)dx$

$du=dx$

$v=-\cos(x)$

Maka

$\begin{aligned}\int x\sin(x)dx & =uv-\int v\mbox{ }du\\
& =-x\cos(x)+\int\cos(x)dx\\
& =-x\cos(x)+\sin(x)+c
\end{aligned}
$

Maka

$\begin{aligned}\int x^{2}\cos(x)dx & =uv-\int vdu\\
& =x^{2}\sin(x)-2\int x\sin(x)dx
\end{aligned}
$

$\int x\sin(x)dx=-x\cos(x)+\sin(x)+c$, jadi

$\int x^{2}\cos(x)dx=x^{2}\sin(x)$$+2x\cos(x)-2\sin(x)+c.$

Ambil,

$u=\sin(\sqrt{x})$

$dv=dx$

$du=\frac{1}{2\sqrt{x}}\cos(\sqrt{x})dx$

$v=x$

Maka

$\begin{aligned}\int\sin(\sqrt{x})dx & =uv-\int vdu\\
& =x\sin(\sqrt{x})-\int\frac{1}{2}\sqrt{x}\cos(\sqrt{x})dx
\end{aligned}
$

Tulis $z=\sqrt{x}$, maka $dz=\frac{1}{2\sqrt{x}}dx$.

Jadi

$\begin{aligned}\int\sin(\sqrt{x})\, dx & =x\sin(\sqrt{x})-\int x\cos(\sqrt{x})\, dz\\
& =x\sin(\sqrt{x})-\int z^{2}\cos(z)\, dz
\end{aligned}
$

Ambil,

$u=x$

$dv=\sin(x)\, dx$

$du=dx$

$v=-\cos(x)$

Maka

$\begin{aligned}\int x\sin(x)\, dx & =uv-\int v\, du\\
& =-x\cos(x)+\int\cos(x)\, dx\\
& =-x\cos(x)+\sin(x)+c
\end{aligned}
$

Maka

$\begin{aligned}\int x^{2}\cos(x)\, dx & =uv-\int v\, du\\
& =x^{2}\sin(x)-2\int x\sin(x)\, dx
\end{aligned}
$

$\int x\sin(x)dx=-x\cos(x)$$+\sin(x)+c$, jadi

$\int x^{2}\cos(x)dx=x^{2}\sin(x)$$+2x\cos(x)-2\sin(x)+c$

Jadi

$\int\sin(\sqrt{x})dx$$=x\sin(\sqrt{x})-x\sin(\sqrt{x})$$-2\sqrt{x}\cos(\sqrt{x})+2\sin(\sqrt{x})+c$

$=-2\sqrt{x}\cos(\sqrt{x})$$+2\sin(\sqrt{x})+c.$

Ambil

$u=\sin(\sqrt{x})$

$dv=dx$

$du=\frac{1}{2\sqrt{x}}\cos(\sqrt{x})dx$

$v=x$

Maka

$\begin{aligned}\int\sin(\sqrt{x})\, dx & =uv-\int v\, du\\
& =x\sin(\sqrt{x})-\int\frac{1}{2}\sqrt{x}\cos(\sqrt{x})dx
\end{aligned}
$

Tulis $z=\sqrt{x}$, maka $dz=\frac{1}{2\sqrt{x}}\, dx$.

Jadi

$\begin{aligned}\int\sin(\sqrt{x})\, dx & =x\sin(\sqrt{x})-\int x\cos(\sqrt{x})\, dz\\
& =x\sin(\sqrt{x})-\int z^{2}\cos(z)\, dz
\end{aligned}
$

Ambil,

$u=x$

$dv=\sin(x)\, dx$

$du=dx$

$v=-\cos(x)$

Maka

$\begin{aligned}\int x\sin(x)\, dx & =uv-\int v\, du\\
& =-x\cos(x)+\int\cos(x)\, dx\\
& =-x\cos(x)+\sin(x)+c
\end{aligned}
$

Maka

$\begin{aligned}\int x^{2}\cos(x)\, dx & =uv-\int v\, du\\
& =x^{2}\sin(x)-2\int x\sin(x)\, dx
\end{aligned}
$

$\int x\sin(x)dx=-x\cos(x)+\sin(x)+c$, jadi

$\int x^{2}\cos(x)dx=x^{2}\sin(x)$$+2x\cos(x)-2\sin(x)+c$

$\begin{aligned}\end{aligned}
$Jadi

$\int\sin(\sqrt{x})dx$$=x\sin(\sqrt{x})-x\sin(\sqrt{x})$$-2\sqrt{x}\cos(\sqrt{x})+2\sin(\sqrt{x})+c$

$=-2\sqrt{x}\cos(\sqrt{x})+2\sin(\sqrt{x})+c$

$\int\sin(\sqrt{4x-2})dx$$=\frac{1}{4}(-2\sqrt{u}\cos(\sqrt{u})+2\sin(\sqrt{u}))+c$

$=-\frac{1}{2}\sqrt{4x-2}\cos(\sqrt{4x-2})$$+\frac{1}{2}\sin(\sqrt{4x-2})+c.$

Ambil;

$u=x$

$dv=\sin(x)\, dx$

$du=dx$

$v=-\cos(x)$

Maka,

$\begin{aligned}\int x\sin(x)\, dx & =uv-\int v\, du\\
& =-x\cos(x)+\int\cos(x)\, dx\\
& =-x\cos(x)+\sin(x)+c
\end{aligned}
$

Maka

$\begin{aligned}\int x^{2}\cos(x)dx & =uv-\int v\, du\\
& =x^{2}\sin(x)-2\int x\sin(x)\, dx
\end{aligned}
$

$\int x\sin(x)\, dx=-x\cos(x)+\sin(x)+c$, jadi

$\int x^{2}\cos(x)dx=x^{2}\sin(x)$$+2x\cos(x)-2\sin(x)+c$

$\int x^{3}\sin(x)dx$$=-x^{3}\cos(x)+3x^{2}\sin(x)$$+6x\cos(x)-6\sin(x)+c.$