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No.
$\displaystyle\int\dfrac{x+2}{\sqrt{x^2+4x-6}}\ dx=$ ....- $\dfrac18\sqrt{x^2+4x-6}+C$
- $\dfrac14\sqrt{x^2+4x-6}+C$
- $\dfrac12\sqrt{x^2+4x-6}+C$
- $\sqrt{x^2+4x-6}+C$
- $2\sqrt{x^2+4x-6}+C$
ALTERNATIF PENYELESAIAN
Misal $u=x^2+4x-6$
\(\begin{aligned} du&=(2x+4)\ dx\\ du&=2(x+2)\ dx\\ (x+2)\ dx&=\dfrac12du \end{aligned}\)
\(\begin{aligned} \displaystyle\int\dfrac{x+2}{\sqrt{x^2+4x-6}}\ dx&=\dfrac12\displaystyle\int\dfrac1{\sqrt{u}}\ du\\ &=\dfrac12\displaystyle\int\dfrac1{u^{\frac12}}\ du\\ &=\dfrac12\displaystyle\int u^{-\frac12}\ du\\ &=\dfrac12\left(2u^{\frac12}\right)+C\\ &=\sqrt{u}+C\\ &=\boxed{\boxed{\sqrt{x^2+4x-6}+C}} \end{aligned}\)
\(\begin{aligned} du&=(2x+4)\ dx\\ du&=2(x+2)\ dx\\ (x+2)\ dx&=\dfrac12du \end{aligned}\)
\(\begin{aligned} \displaystyle\int\dfrac{x+2}{\sqrt{x^2+4x-6}}\ dx&=\dfrac12\displaystyle\int\dfrac1{\sqrt{u}}\ du\\ &=\dfrac12\displaystyle\int\dfrac1{u^{\frac12}}\ du\\ &=\dfrac12\displaystyle\int u^{-\frac12}\ du\\ &=\dfrac12\left(2u^{\frac12}\right)+C\\ &=\sqrt{u}+C\\ &=\boxed{\boxed{\sqrt{x^2+4x-6}+C}} \end{aligned}\)
No.
Hasil dari $\displaystyle\int-720x^2\sqrt{9-4x^3}\ dx=$- ${80\left(9-4x^3\right)\sqrt{9-4x^3}+C}$
- ${60\left(9-4x^3\right)\sqrt{9-4x^3}+C}$
- ${40\left(9-4x^3\right)\sqrt{9-4x^3}+C}$
- ${20\left(9-4x^3\right)\sqrt{9-4x^3}+C}$
- ${10\left(9-4x^3\right)\sqrt{9-4x^3}+C}$
ALTERNATIF PENYELESAIAN
Misal $u=9-4x^3$
\(\begin{aligned} du&=-12x^2\ dx \end{aligned}\)
\(\begin{aligned} \displaystyle\int-720x^2\sqrt{9-4x^3}\ dx&=\displaystyle\int60\left(-12x^2\right)\sqrt{9-4x^3}\ dx\\ &=\displaystyle\int60\sqrt{u}\ du\\ &=\displaystyle\int60u^{\frac12}\ du\\ &=60\cdot\dfrac23u^{\frac32}+C\\ &=40u\sqrt{u}+C\\ &=\boxed{\boxed{40\left(9-4x^3\right)\sqrt{9-4x^3}+C}} \end{aligned}\)
\(\begin{aligned} du&=-12x^2\ dx \end{aligned}\)
\(\begin{aligned} \displaystyle\int-720x^2\sqrt{9-4x^3}\ dx&=\displaystyle\int60\left(-12x^2\right)\sqrt{9-4x^3}\ dx\\ &=\displaystyle\int60\sqrt{u}\ du\\ &=\displaystyle\int60u^{\frac12}\ du\\ &=60\cdot\dfrac23u^{\frac32}+C\\ &=40u\sqrt{u}+C\\ &=\boxed{\boxed{40\left(9-4x^3\right)\sqrt{9-4x^3}+C}} \end{aligned}\)
No.
$\displaystyle\int\left(\dfrac1{\sqrt{x}}-\sqrt{x}\right)^2\ dx$ adalah- ${\dfrac12x^2-2x+\ln x+C}$
- ${x^2-2+\dfrac1x+C}$
- ${x^2-2x+x^{-1}+C}$
- ${\dfrac12x^2-2x+x^{-1}+C}$
- ${\dfrac12x^2-2+\dfrac1x+C}$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\int\left(\dfrac1{\sqrt{x}}-\sqrt{x}\right)^2\ dx&=\displaystyle\int\left(\left(\dfrac1{\sqrt{x}}\right)^2-2\cdot\dfrac1{\sqrt{x}}\cdot\sqrt{x}+\left(\sqrt{x}\right)^2\right)\ dx\\
&=\displaystyle\int\left(\dfrac1x-2+x\right)\ dx\\
&=\ln x-2x+\dfrac12x^2+C\\
&=\boxed{\boxed{\dfrac12x^2-2x+\ln x+C}}
\end{aligned}\)
No.
$\displaystyle\int2\sqrt{x}\ dx=$ ...- $\sqrt{x}+C$
- $x\sqrt{x}+C$
- $2\sqrt{x}+C$
- $\dfrac43x\sqrt{x}+C$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\int2\sqrt{x}\ dx&=\displaystyle\int2x^{\frac12}\ dx\\
&=2\cdot\dfrac23x^{\frac32}+C\\
&=\boxed{\boxed{\dfrac43x\sqrt{x}+C}}
\end{aligned}\)
No.
\[\int\frac{e^{8+\frac1x}}{x^2}\ dx=....\]- $\dfrac{e^{8+\frac1x}}{x^2}+C$
- $-\dfrac{e^{8+\frac1x}}{x^2}+C$
- $e^{8+\frac1x}+C$
- $-e^{8+\frac1x}+C$
ALTERNATIF PENYELESAIAN
Misalkan $u=8+\dfrac1x$.
$du=-\dfrac1{x^2}\ dx$ \begin{aligned} \int\frac{e^{8+\frac1x}}{x^2}\ dx&=-\int\left(e^{8+\frac1x}\right)\left(-\frac1{x^2}\right)\ dx\\ &=-\int e^u\ du\\ &=-e^u+C\\ &=\boxed{\boxed{-e^{8+\frac1x}+C}} \end{aligned}
$du=-\dfrac1{x^2}\ dx$ \begin{aligned} \int\frac{e^{8+\frac1x}}{x^2}\ dx&=-\int\left(e^{8+\frac1x}\right)\left(-\frac1{x^2}\right)\ dx\\ &=-\int e^u\ du\\ &=-e^u+C\\ &=\boxed{\boxed{-e^{8+\frac1x}+C}} \end{aligned}
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