SBMPTN Zone : Integral Tak Tentu

Berikut ini adalah kumpulan soal mengenai integral tak tentu tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN

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

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.

No. 3

\displaystyle\int\dfrac{x^2-2x-1}{x^2-1}\ dx=

1. x-\ln\left|x^2-1\right|+c
2. x-2\ln\left|x^2-1\right|+c
3. 2x-2\ln\left|3x-1\right|+c

4. 2x-4\ln\left|x^2-1\right|+c
5. 2x-4\ln\left|3x-1\right|+c

SKMP September 2020

\(\eqalign{ \displaystyle\int\dfrac{x^2-2x-1}{x^2-1}\ dx&=\displaystyle\int\dfrac{x^2-1-2x}{x^2-1}\ dx\\ &=\displaystyle\int\left(1-\dfrac{2x}{x^2-1}\right)\ dx\\ &=\displaystyle\int dx-\displaystyle\int\dfrac{2x}{x^2-1}\ dx\\ &=x-\displaystyle\int\dfrac1{x^2-1}\ d\left(x^2-1\right)\\ &=\boxed{\boxed{x-\ln\left|x^2-1\right|+c}} }\)

Jadi, \displaystyle\int\dfrac{x^2-2x-1}{x^2-1}\ dx=x-\ln\left|x^2-1\right|+c.
JAWAB: A

No. 4

\displaystyle\int\dfrac{dx}{x-\sqrt{x}}

Channel Telegram

\displaystyle\int\dfrac{dx}{x-\sqrt{x}}=\displaystyle\int\dfrac{dx}{\sqrt{x}\left(\sqrt{x}-1\right)}

Misal u=\sqrt{x}-1
\(\eqalign{ du&=\dfrac{dx}{2\sqrt{x}}\\ \dfrac{dx}{\sqrt{x}}&=2du }\)

\(\eqalign{ \displaystyle\int\dfrac{dx}{\sqrt{x}\left(\sqrt{x}-1\right)}&=2\displaystyle\int\dfrac{du}u\\ &=2\ln|u|+C\\ &=\boxed{\boxed{2\ln\left|\sqrt{x}-1\right|+C}} }\)

Jadi, \displaystyle\int\dfrac{dx}{x-\sqrt{x}}=2\ln\left|\sqrt{x}-1\right|+C.