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

$\begin{array}{rl}{\displaystyle \int g(x)dx}& =3\sqrt{f(x)}+c\\ g(x)& ={\displaystyle \frac{d(3\sqrt{f(x)}+c)}{dx}}\\ & ={\displaystyle \frac{d(3{(f(x))}^{\frac{1}{2}}+c)}{dx}}\\ & =3\cdot {\displaystyle \frac{1}{2}}{(f(x))}^{-\frac{1}{2}}{f}^{\prime }(x)\\ & ={\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{1}{{(f(x))}^{\frac{1}{2}}}}\cdot f^{\prime }(x)\\ & ={\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{1}{\sqrt{f(x)}}}\cdot f^{\prime }(x)\\ & ={\displaystyle \frac{3f^{\prime }(x)}{2\sqrt{f(x)}}}\\ g(1)& ={\displaystyle \frac{3f^{\prime }(1)}{2\sqrt{f(1)}}}\\ & ={\displaystyle \frac{3(9)}{2\sqrt{9}}}\\ & ={\displaystyle \frac{27}{2(3)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{9}{2}}}}}}\end{array}$

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

No. 2

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

Misal
$\begin{array}{rl}u& =1+e^x\\ du& =e^{x}dx\\ du& =(u-1)dx\\ dx& ={\displaystyle \frac{1}{u-1}}du\end{array}$

$\begin{array}{rl}{\displaystyle \int {\displaystyle \frac{1}{1+e^x}}dx}& ={\displaystyle \int {\displaystyle \frac{1}{u}}\cdot {\displaystyle \frac{1}{u-1}}du}\\ & ={\displaystyle \int ({\displaystyle \frac{-1}{u}}+{\displaystyle \frac{1}{u-1}})du}\\ & =-\ln |u|+\ln |u-1|+C\\ & =-\ln |1+e^x|+\ln |1+e^x-1|+C\\ & =-\ln |1+e^x|+\ln |e^x|+C\\ & =-\ln |1+e^x|+x+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle x-\ln |1+e^x|+C}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \int {\displaystyle \frac{x^2-2x-1}{x^2-1}}dx}& ={\displaystyle \int {\displaystyle \frac{x^2-1-2x}{x^2-1}}dx}\\ & ={\displaystyle \int (1-{\displaystyle \frac{2x}{x^2-1}})dx}\\ & ={\displaystyle \int dx-{\displaystyle \int {\displaystyle \frac{2x}{x^2-1}}dx}}\\ & =x-{\displaystyle \int {\displaystyle \frac{1}{x^2-1}}d(x^2-1)}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle x-\ln |x^2-1|+c}}}}\end{array}$

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
$\begin{array}{rl}du& ={\displaystyle \frac{dx}{2\sqrt{x}}}\\ {\displaystyle \frac{dx}{\sqrt{x}}}& =2du\end{array}$

$\begin{array}{rl}{\displaystyle \int {\displaystyle \frac{dx}{\sqrt{x}(\sqrt{x}-1)}}}& =2{\displaystyle \int {\displaystyle \frac{du}{u}}}\\ & =2\ln |u|+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2\ln |\sqrt{x}-1|+C}}}}\end{array}$

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