Exercise Zone : Integral Tak Tentu

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

Tipe:

Standar SBMPTN HOTS

No. 1

${\displaystyle \int {\displaystyle \frac{x+2}{\sqrt{x^2+4x-6}}}dx=}$ ....

1. ${\displaystyle \frac{1}{8}}\sqrt{x^2+4x-6}+C$
2. ${\displaystyle \frac{1}{4}}\sqrt{x^2+4x-6}+C$
3. ${\displaystyle \frac{1}{2}}\sqrt{x^2+4x-6}+C$

4. $\sqrt{x^2+4x-6}+C$
5. $2\sqrt{x^2+4x-6}+C$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

Misal $u=x^2+4x-6$
$\begin{array}{rl}du& =(2x+4)dx\\ du& =2(x+2)dx\\ (x+2)dx& ={\displaystyle \frac{1}{2}}du\end{array}$

$\begin{array}{rl}{\displaystyle \int {\displaystyle \frac{x+2}{\sqrt{x^2+4x-6}}}dx}& ={\displaystyle \frac{1}{2}}{\displaystyle \int {\displaystyle \frac{1}{\sqrt{u}}}du}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle \int {\displaystyle \frac{1}{u^{\frac{1}{2}}}}du}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle \int u^{-\frac{1}{2}}du}\\ & ={\displaystyle \frac{1}{2}}\left(2u^{\frac{1}{2}}\right)+C\\ & =\sqrt{u}+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{x^2+4x-6}+C}}}}\end{array}$

Jadi, ${\displaystyle \int {\displaystyle \frac{x+2}{\sqrt{x^2+4x-6}}}dx=\sqrt{x^2+4x-6}+C}$.
JAWAB: D

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No. 2

Hasil dari ${\displaystyle \int -720x^2\sqrt{9-4x^3}dx=}$

1. $80(9-4x^3)\sqrt{9-4x^3}+C$
2. $60(9-4x^3)\sqrt{9-4x^3}+C$
3. $40(9-4x^3)\sqrt{9-4x^3}+C$

4. $20(9-4x^3)\sqrt{9-4x^3}+C$
5. $10(9-4x^3)\sqrt{9-4x^3}+C$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

Misal $u=9-4x^3$
$\begin{array}{rl}du& =-12x^{2}dx\end{array}$

$\begin{array}{rl}{\displaystyle \int -720x^2\sqrt{9-4x^3}dx}& ={\displaystyle \int 60(-12x^2)\sqrt{9-4x^3}dx}\\ & ={\displaystyle \int 60\sqrt{u}du}\\ & ={\displaystyle \int 60u^{\frac{1}{2}}du}\\ & =60\cdot {\displaystyle \frac{2}{3}}{u}^{\frac{3}{2}}+C\\ & =40u\sqrt{u}+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 40(9-4x^3)\sqrt{9-4x^3}+C}}}}\end{array}$

Jadi, ${\displaystyle \int -720x^2\sqrt{9-4x^3}dx=40(9-4x^3)\sqrt{9-4x^3}+C}$.
JAWAB: C

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No. 3

${\displaystyle \int {({\displaystyle \frac{1}{\sqrt{x}}}-\sqrt{x})}^{2}dx}$ adalah

1. ${\displaystyle \frac{1}{2}}{x}^2-2x+\ln x+C$
2. $x^2-2+{\displaystyle \frac{1}{x}}+C$
3. $x^2-2x+x^{-1}+C$

4. ${\displaystyle \frac{1}{2}}{x}^2-2x+x^{-1}+C$
5. ${\displaystyle \frac{1}{2}}{x}^2-2+{\displaystyle \frac{1}{x}}+C$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \int {({\displaystyle \frac{1}{\sqrt{x}}}-\sqrt{x})}^{2}dx}& ={\displaystyle \int ({\left({\displaystyle \frac{1}{\sqrt{x}}}\right)}^2-2\cdot {\displaystyle \frac{1}{\sqrt{x}}}\cdot \sqrt{x}+{\left(\sqrt{x}\right)}^2)dx}\\ & ={\displaystyle \int ({\displaystyle \frac{1}{x}}-2+x)dx}\\ & =\ln x-2x+{\displaystyle \frac{1}{2}}{x}^2+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}{x}^2-2x+\ln x+C}}}}\end{array}$

Jadi, ${\displaystyle \int {({\displaystyle \frac{1}{\sqrt{x}}}-\sqrt{x})}^{2}dx={\displaystyle \frac{1}{2}}{x}^2-2x+\ln x+C}$.
JAWAB: A

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No. 4

${\displaystyle \int 2\sqrt{x}dx=}$ ...

1. $\sqrt{x}+C$
2. $x\sqrt{x}+C$

3. $2\sqrt{x}+C$
4. ${\displaystyle \frac{4}{3}}x\sqrt{x}+C$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \int 2\sqrt{x}dx}& ={\displaystyle \int 2x^{\frac{1}{2}}dx}\\ & =2\cdot {\displaystyle \frac{2}{3}}{x}^{\frac{3}{2}}+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}x\sqrt{x}+C}}}}\end{array}$

Jadi, ${\displaystyle \int 2\sqrt{x}dx={\displaystyle \frac{4}{3}}x\sqrt{x}+C}$.
JAWAB: D

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No. 5

$$\int \frac{e^{8+\frac{1}{x}}}{x^2}dx=....$$

1. ${\displaystyle \frac{e^{8+\frac{1}{x}}}{x^2}}+C$
2. $-{\displaystyle \frac{e^{8+\frac{1}{x}}}{x^2}}+C$

3. $e^{8+\frac{1}{x}}+C$
4. $-e^{8+\frac{1}{x}}+C$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

Misalkan $u=8+{\displaystyle \frac{1}{x}}$.
$du=-{\displaystyle \frac{1}{x^2}}dx$ $$\begin{array}{rl}\int \frac{e^{8+\frac{1}{x}}}{x^2}dx& =-\int \left(e^{8+\frac{1}{x}}\right)(-\frac{1}{x^2})dx\\ & =-\int e^{u}du\\ & =-e^u+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -e^{8+\frac{1}{x}}+C}}}}\end{array}$$

Jadi, $$\int \frac{e^{8+\frac{1}{x}}}{x^2}dx=-e^{8+\frac{1}{x}}+C$$JAWAB: D

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