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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