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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