Exercise 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

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

1. ${\dfrac18\sqrt{x^2+4x-6}+C}$
2. ${\dfrac14\sqrt{x^2+4x-6}+C}$
3. ${\dfrac12\sqrt{x^2+4x-6}+C}$

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

Misal $u=x^2+4x-6$
$\begin{aligned}du&=(2x+4)\ dx\\du&=2(x+2)\ dx\\(x+2)\ dx&=\dfrac12du\end{aligned}$

$\begin{aligned}\displaystyle\int\dfrac{x+2}{\sqrt{x^2+4x-6}}\ dx&=\dfrac12\displaystyle\int\dfrac1{\sqrt{u}}\ du\\&=\dfrac12\displaystyle\int\dfrac1{u^{\frac12}}\ du\\&=\dfrac12\displaystyle\int u^{-\frac12}\ du\\&=\dfrac12\left(2u^{\frac12}\right)+C\\&=\sqrt{u}+C\\&=\boxed{\boxed{\sqrt{x^2+4x-6}+C}}\end{aligned}$

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

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No. 2

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

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

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

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

$\begin{aligned}\displaystyle\int-720x^2\sqrt{9-4x^3}\ dx&=\displaystyle\int60\left(-12x^2\right)\sqrt{9-4x^3}\ dx\\&=\displaystyle\int60\sqrt{u}\ du\\&=\displaystyle\int60u^{\frac12}\ du\\&=60\cdot\dfrac23u^{\frac32}+C\\&=40u\sqrt{u}+C\\&=\boxed{\boxed{40\left(9-4x^3\right)\sqrt{9-4x^3}+C}}\end{aligned}$

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

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No. 3

Hasil dari $\displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx=$

1. $-\dfrac{120}{147}$
2. $-\dfrac{157}{120}$
3. $-\dfrac{147}{120}$

4. $\dfrac{147}{120}$
5. $\dfrac{120}{147}$

CARA 1: SUBSTITUSI

Misal $u=1+x\rightarrow x=u-1$
$du=dx$

$\begin{aligned}\displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx&=\displaystyle\intop_{-1}^03(u-1)\sqrt[7]{u}\ du\\&=\displaystyle\intop_{-1}^03(u-1)u^{\frac17}\ du\\&=\displaystyle\intop_{-1}^0\left(3u^{\frac87}-3u^{\frac17}\right)\ du\\&=\left[3\cdot\dfrac7{15}u^{\frac{15}7}-3\cdot\dfrac78u^{\frac87}\right]_{-1}^0\\&=\left[\dfrac75(1+x)^{\frac{15}7}-\dfrac{21}8(1+x)^{\frac87}\right]_{-1}^0\\&=\left[\dfrac75(1+0)^{\frac{15}7}-\dfrac{21}8(1+0)^{\frac87}\right]-\left[\dfrac75(1+(-1))^{\frac{15}7}-\dfrac{21}8(1+(-1))^{\frac87}\right]\\&=\left[\dfrac75(1)^{\frac{15}7}-\dfrac{21}8(1)^{\frac87}\right]-\left[\dfrac75(0)^{\frac{15}7}-\dfrac{21}8(0)^{\frac87}\right]\\&=\left[\dfrac75(1)-\dfrac{21}8(1)\right]-\left[\dfrac75(0)-\dfrac{21}8(0)\right]\\&=\left[\dfrac75-\dfrac{21}8\right]-\left[0-0\right]\\&=-\dfrac{49}{40}\\&=\boxed{\boxed{-\dfrac{147}{120}}}\end{aligned}$

CARA 2: PARSIAL

$u$	$dv$
$3x$	$\sqrt[7]{1+x}=(1+x)^{\frac17}$
$3$	$\dfrac78(1+x)^{\frac87}$
$0$	$\dfrac78\cdot\dfrac7{15}(1+x)^{\frac{15}7}=\dfrac{49}{120}(1+x)^{\frac{15}7}$

$\begin{aligned}\displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx&=\left[3x\cdot\dfrac78(1+x)^{\frac87}-3\cdot\dfrac{49}{120}(1+x)^{\frac{15}7}\right]_{-1}^0\\&=\left[\dfrac{21}8x(1+x)^{\frac87}-\cdot\dfrac{147}{120}(1+x)^{\frac{15}7}\right]_{-1}^0\\&=\left[\dfrac{21}8(0)(1+0)^{\frac87}-\cdot\dfrac{147}{120}(1+0)^{\frac{15}7}\right]-\left[\dfrac{21}8(-1)(1+(-1))^{\frac87}-\cdot\dfrac{147}{120}(1+(-1))^{\frac{15}7}\right]\\&=\left[0-\cdot\dfrac{147}{120}(1)^{\frac{15}7}\right]-\left[-\dfrac{21}8(0)^{\frac87}-\cdot\dfrac{147}{120}(0)^{\frac{15}7}\right]\\&=\left[-\cdot\dfrac{147}{120}(1)\right]-0\\&=\boxed{\boxed{-\dfrac{147}{120}}}\end{aligned}$

Jadi, $\displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx=-\dfrac{147}{120}$.
JAWAB: C

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No. 4

$\displaystyle\int\left(\dfrac1{\sqrt{x}}-\sqrt{x}\right)^2\ dx$ adalah

1. ${\dfrac12x^2-2x+\ln x+C}$
2. ${x^2-2+\dfrac1x+C}$
3. ${x^2-2x+x^{-1}+C}$

4. ${\dfrac12x^2-2x+x^{-1}+C}$
5. ${\dfrac12x^2-2+\dfrac1x+C}$

$\begin{aligned}\displaystyle\int\left(\dfrac1{\sqrt{x}}-\sqrt{x}\right)^2\ dx&=\displaystyle\int\left(\left(\dfrac1{\sqrt{x}}\right)^2-2\cdot\dfrac1{\sqrt{x}}\cdot\sqrt{x}+\left(\sqrt{x}\right)^2\right)\ dx\\&=\displaystyle\int\left(\dfrac1x-2+x\right)\ dx\\&=\ln x-2x+\dfrac12x^2+C\\&=\boxed{\boxed{\dfrac12x^2-2x+\ln x+C}}\end{aligned}$

Jadi, $\displaystyle\int\left(\dfrac1{\sqrt{x}}-\sqrt{x}\right)^2\ dx=\dfrac12x^2-2x+\ln x+C$.
JAWAB: A

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No. 5

$\displaystyle\int2\sqrt{x}\ dx=$

$\begin{aligned}\displaystyle\int2\sqrt{x}\ dx&=\displaystyle\int2x^{\frac12}\ dx\\&=2\cdot\dfrac23x^{\frac32}+C\\&=\boxed{\boxed{\dfrac43x\sqrt{x}+C}}\end{aligned}$

Jadi, $\displaystyle\int2\sqrt{x}\ dx=\dfrac43x\sqrt{x}+C$.

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