SNBT Zone : Integral Tak Tentu
Table of Contents
Tipe:
No.
Jika \(\displaystyle\int g(x)\ dx=3\sqrt{f(x)}+c\) dan f(1) = f'(1) = 9 maka g(1) =- 1
- 9
- 3
- \(\dfrac32\)
- \(\dfrac92\)
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\int g(x)\ dx&=3\sqrt{f(x)}+c\\
g(x)&=\dfrac{d\left(3\sqrt{f(x)}+c\right)}{dx}\\
&=\dfrac{d\left(3\left(f(x)\right)^{\frac12}+c\right)}{dx}\\
&=3\cdot\dfrac12\left(f(x)\right)^{-\frac12}f'(x)\\[3.5pt]
&=\dfrac32\cdot\dfrac1{\left(f(x)\right)^{\frac12}}\cdot f'(x)\\[3.5pt]
&=\dfrac32\cdot\dfrac1{\sqrt{f(x)}}\cdot f'(x)\\[3.5pt]
&=\dfrac{3f'(x)}{2\sqrt{f(x)}}\\[3.5pt]
g(1)&=\dfrac{3f'(1)}{2\sqrt{f(1)}}\\[3.5pt]
&=\dfrac{3(9)}{2\sqrt9}\\[3.5pt]
&=\dfrac{27}{2(3)}\\
&=\boxed{\boxed{\dfrac92}}
\end{aligned}
Jadi, \(g(1)=\dfrac92\).
JAWAB: E
JAWAB: E
No.
\[\displaystyle\int\dfrac1{1+e^x}\ dx=\]ALTERNATIF PENYELESAIAN
\(\begin{aligned}
u&=1+e^x\\
du&=e^x\ dx\\
du&=(u-1)\ dx\\
dx&=\dfrac1{u-1}\ du
\end{aligned}\)
\(\begin{aligned} \displaystyle\int\dfrac1{1+e^x}\ dx&=\displaystyle\int\dfrac1u\cdot\dfrac1{u-1}\ du\\ &=\displaystyle\int\left(\dfrac{-1}u+\dfrac1{u-1}\right)\ du\\ &=-\ln|u|+\ln|u-1|+C\\ &=-\ln|1+e^x|+\ln|1+e^x-1|+C\\ &=-\ln|1+e^x|+\ln|e^x|+C\\ &=-\ln|1+e^x|+x+C\\ &=\boxed{\boxed{x-\ln|1+e^x|+C}} \end{aligned}\)
\(\begin{aligned} \displaystyle\int\dfrac1{1+e^x}\ dx&=\displaystyle\int\dfrac1u\cdot\dfrac1{u-1}\ du\\ &=\displaystyle\int\left(\dfrac{-1}u+\dfrac1{u-1}\right)\ du\\ &=-\ln|u|+\ln|u-1|+C\\ &=-\ln|1+e^x|+\ln|1+e^x-1|+C\\ &=-\ln|1+e^x|+\ln|e^x|+C\\ &=-\ln|1+e^x|+x+C\\ &=\boxed{\boxed{x-\ln|1+e^x|+C}} \end{aligned}\)
Jadi, \(\displaystyle\int\dfrac1{1+e^x}\ dx=x-\ln|1+e^x|+C\).
No.
\[\displaystyle\int\dfrac{x^2-2x-1}{x^2-1}\ dx=\]- x − ln |x2 − 1| + C
- x − 2 ln |x2 − 1| + C
- 2x − 2 ln |3x − 1| + C
- 2x − 4 ln |x2 − 1| + C
- 2x − 4 ln |3x − 1| + C
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\int\dfrac{x^2-2x-1}{x^2-1}\ dx&=\displaystyle\int\dfrac{x^2-1-2x}{x^2-1}\ dx\\
&=\displaystyle\int\left(1-\dfrac{2x}{x^2-1}\right)\ dx\\
&=\displaystyle\int dx-\displaystyle\int\dfrac{2x}{x^2-1}\ dx\\
&=x-\displaystyle\int\dfrac1{x^2-1}\ d\left(x^2-1\right)\\
&=\boxed{\boxed{x-\ln\left|x^2-1\right|+c}}
\end{aligned}
Jadi, \(\displaystyle\int\dfrac{x^2-2x-1}{x^2-1}\ dx=x-\ln\left|x^2-1\right|+C\).
JAWAB: A
JAWAB: A
No.
\[\displaystyle\int\dfrac{dx}{x-\sqrt{x}}= ....\]ALTERNATIF PENYELESAIAN
\(\displaystyle\int\dfrac{dx}{x-\sqrt{x}}=\displaystyle\int\dfrac{dx}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
Misal \(u=\sqrt{x}-1\) \begin{aligned} du&=\dfrac{dx}{2\sqrt{x}}\\ \dfrac{dx}{\sqrt{x}}&=2du \end{aligned} \begin{aligned} \displaystyle\int\dfrac{dx}{\sqrt{x}\left(\sqrt{x}-1\right)}&=2\displaystyle\int\dfrac{du}u\\ &=2\ln|u|+C\\ &=\boxed{\boxed{2\ln\left|\sqrt{x}-1\right|+C}} \end{aligned}
Misal \(u=\sqrt{x}-1\) \begin{aligned} du&=\dfrac{dx}{2\sqrt{x}}\\ \dfrac{dx}{\sqrt{x}}&=2du \end{aligned} \begin{aligned} \displaystyle\int\dfrac{dx}{\sqrt{x}\left(\sqrt{x}-1\right)}&=2\displaystyle\int\dfrac{du}u\\ &=2\ln|u|+C\\ &=\boxed{\boxed{2\ln\left|\sqrt{x}-1\right|+C}} \end{aligned}
Jadi, \(\displaystyle\int\dfrac{dx}{x-\sqrt{x}}=2\ln\left|\sqrt{x}-1\right|+C\).
No.
\(\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}+3x^2\sqrt{x-1}\ dx\) = ....ALTERNATIF PENYELESAIAN
\(\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}+3x^2\sqrt{x-1}\ dx=\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}\ dx+\displaystyle\int3x^2\sqrt{x-1}\ dx\)
\(\begin{aligned} \displaystyle\int u\ dv&=uv-\displaystyle\int v\ du\\ \displaystyle\int 3x^2\sqrt{x-1}\ dx&=\sqrt{x-1}\cdot x^3-\displaystyle\int x^3\cdot\dfrac1{2\sqrt{x-1}}\ dx+C\\ &=x^3\sqrt{x-1}-\displaystyle\int \dfrac{x^3}{2\sqrt{x-1}}\ dx+C \end{aligned}\)
\(\begin{aligned} \displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}+3x^2\sqrt{x-1}\ dx&=\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}\ dx+\displaystyle\int3x^2\sqrt{x-1}\ dx\\ &=\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}\ dx+x^3\sqrt{x-1}-\displaystyle\int \dfrac{x^3}{2\sqrt{x-1}}\ dx+C\\ &=\boxed{\boxed{x^3\sqrt{x-1}+C}} \end{aligned}\)
\(u=\sqrt{x-1}\)
\(du=\dfrac1{2\sqrt{x-1}}\ dx\)
\(du=\dfrac1{2\sqrt{x-1}}\ dx\)
dv = 3x2
v = x3 + C
v = x3 + C
\(\begin{aligned} \displaystyle\int u\ dv&=uv-\displaystyle\int v\ du\\ \displaystyle\int 3x^2\sqrt{x-1}\ dx&=\sqrt{x-1}\cdot x^3-\displaystyle\int x^3\cdot\dfrac1{2\sqrt{x-1}}\ dx+C\\ &=x^3\sqrt{x-1}-\displaystyle\int \dfrac{x^3}{2\sqrt{x-1}}\ dx+C \end{aligned}\)
\(\begin{aligned} \displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}+3x^2\sqrt{x-1}\ dx&=\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}\ dx+\displaystyle\int3x^2\sqrt{x-1}\ dx\\ &=\displaystyle\int\dfrac{x^3}{2\sqrt{x-1}}\ dx+x^3\sqrt{x-1}-\displaystyle\int \dfrac{x^3}{2\sqrt{x-1}}\ dx+C\\ &=\boxed{\boxed{x^3\sqrt{x-1}+C}} \end{aligned}\)
Jadi,
JAWAB:
JAWAB:
Post a Comment