SBMPTN Zone : Integral Tak Tentu

Berikut ini adalah kumpulan soal mengenai Integral Tak Tentu tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Teirma kasih.

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No. 1

Jika \displaystyle\int g(x)\ dx=3\sqrt{f(x)}+c dan f(1)=f'(1)=9 maka g(1)=

1. 1
2. 9
3. 3

4. \dfrac32
5. \dfrac92

\(\begin{aligned} \displaystyle\int g(x)\ dx&=3\sqrt{f(x)}+c\\ g(x)&=\dfrac{d\left(3\sqrt{f(x)}+c\right)}{dx}\\ &=\dfrac{d\left(3\left(f(x)\right)^{\frac12}+c\right)}{dx}\\ &=3\cdot\dfrac12\left(f(x)\right)^{-\frac12}f'(x)\\[10pt] &=\dfrac32\cdot\dfrac1{\left(f(x)\right)^{\frac12}}\cdot f'(x)\\[10pt] &=\dfrac32\cdot\dfrac1{\sqrt{f(x)}}\cdot f'(x)\\[10pt] &=\dfrac{3f'(x)}{2\sqrt{f(x)}}\\[10pt] g(1)&=\dfrac{3f'(1)}{2\sqrt{f(1)}}\\[10pt] &=\dfrac{3(9)}{2\sqrt9}\\[10pt] &=\dfrac{27}{2(3)}\\ &=\boxed{\boxed{\dfrac92}} \end{aligned}\)

Jadi, g(1)=\dfrac92.
JAWAB: E

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No. 2

\displaystyle\int\dfrac1{1+e^x}\ dx=

Misal
\(\begin{aligned} u&=1+e^x\\ du&=e^x\ dx\\ du&=(u-1)\ dx\\ dx&=\dfrac1{u-1}\ du \end{aligned}\)

\(\begin{aligned} \displaystyle\int\dfrac1{1+e^x}\ dx&=\displaystyle\int\dfrac1u\cdot\dfrac1{u-1}\ du\\ &=\displaystyle\int\left(\dfrac{-1}u+\dfrac1{u-1}\right)\ du\\ &=-\ln|u|+\ln|u-1|+C\\ &=-\ln|1+e^x|+\ln|1+e^x-1|+C\\ &=-\ln|1+e^x|+\ln|e^x|+C\\ &=-\ln|1+e^x|+x+C\\ &=\boxed{\boxed{x-\ln|1+e^x|+C}} \end{aligned}\)

Jadi, \displaystyle\int\dfrac1{1+e^x}\ dx=x-\ln|1+e^x|+C.

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