Please bantuin lahh.... Question from @sintyasuherlan

October 2018 1 49 Report

Please bantuin lahh

Takamori37 Nomot 7.
Bag a.
Dengan:
u = 5-x²
du = -2x dx, menjadi x dx = -1/2 du
Diperoleh:
$$\begin{align}\int\frac{x}{\sqrt{5-x^2}}\, dx&=\int-\frac12\times\frac{1}{\sqrt{u}}\, du \\ &=-\frac12\int u^{-1/2}\, du \\ &=-\frac12\times\frac{u^{1/2}}{1/2}+C \\ &=-\sqrt{u}+C \\ &=-\sqrt{5-x^2}+C\end{align}$

Bag b.
Dengan:
u = 5+x³
du = 3x² dx, menjadi x² dx = 1/3 du
Diperoleh:
$$\begin{align}\int\frac{x^2}{(5+x^3)}\,
 dx&=\int\frac13\times\frac{1}{u^2}\, du \\ &=\frac13\int
 u^{-2}\, du \\ &=\frac13\times\frac{u^{-1}}{-1}+C \\ 
&=\frac{1}{3u}+C \\ &=\frac{1}{3(5+x^3)}+C\end{align}$

Bag c.
Dengan:
u = 4-9x²
du = -18x dx, menjadi x dx = -1/18 du
Diperoleh:
$$\begin{align}\int\frac{x}{\sqrt{4-9x^2}}\,
 dx&=\int-\frac1{18}\times\frac{1}{\sqrt{u}}\, du \\ &=-\frac1{18}\int
 u^{-1/2}\, du \\ &=-\frac1{18}\times\frac{u^{1/2}}{1/2}+C \\ 
&=-\frac19\sqrt{u}+C \\ &=-\frac19\sqrt{4-9x^2}+C\end{align}$

Bag d.
Dengan:
$\boxed{\int\frac{dx}{(x+a)^2+b^2}=\frac1b\tan^{-1}\frac{x+a}{b}+C}$
Diperoleh:
$$\begin{align}\int\frac{dx}{x^2+6x+13}&=\int\frac{dx}{(x^2+6x+9)+4} \\ &=\int\frac{dx}{(x+3)^2+2^2} \\ &=\frac12\tan^{-1}\left[\frac{x+3}{2}\right]+C\end{align}$

Nomor 9.
Bag a.
Dengan:
u = 1 + √x
du = 1/[2√x] dx, menjadi 1/√x dx = 2 du
Diperoleh:
$$\begin{align}\int\frac{(1+\sqrt x)^2}{\sqrt x}\, dx&=\int2u^2\, du \\ &=2\times\frac{u^3}{3}+C \\ &=\frac23(1+\sqrt{x})^3+C\end{align}$

Bag b.
$$\begin{align}\int\frac{(x+1)(x-2)}{\sqrt x}\, dx&=\int\frac{x^2-x-2}{\sqrt x}\, dx \\ &=x^{3/2}-x^{1/2}-2x^{-1/2}\, dx \\ &=\frac{x^{5/2}}{5/2}-\frac{x^{3/2}}{3/2}-\frac{2x^{1/2}}{1/2}+C \\ &=\frac25x^{5/2}-\frac23x^{3/2}-4x^{1/2}+C \\ &=\sqrt x \left[\frac25x^2-\frac23x-4\right]+C\end{align}$

Bag c.
Dengan:
u = 1-2x²
du = -4x dx, menjadi 5x dx = -5/4 du
Diperoleh:
$$\begin{align}\int 5x\sqrt{1-2x^2}\, dx&=\int-\frac54\sqrt{u}\, du \\ &=-\frac54\times\int u^{1/2}\, du \\ &=-\frac54\times\frac{u^{3/2}}{3/2}+C \\ &=-\frac56u^{3/2}+C \\ &=-\frac56(1-2x^2)^{3/2}+C\end{align}$

Bag d.
Dengan:
u = 1+x⁴
du = 4x³ dx, menjadi 5x³ dx = 5/4 du
Diperoleh:
$$\begin{align}\int\frac{5x^3}{\sqrt{1+x^4}}\,
 dx&=\int\frac54\times\frac{1}{\sqrt{u}}\, du \\ &=\frac54\int
 u^{-1/2}\, du \\ &=\frac54\times\frac{u^{1/2}}{1/2}+C \\ 
&=\frac52\sqrt{u}+C \\ &=\frac52\sqrt{1+x^4}+C\end{align}$

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