MATEMATIKA INDONESIA: Integral Tak Tentu 2

Soal-soal Integral tak tentu
$1. \int (x^{5}-6x^{2}-4x+3)dx = \frac{1}{5+1}x^{5+1}-\frac{6}{2+1}x^{2+1}-\frac{4}{1+1}x^{1+1}+3x+c$
$=\frac{1}{6}x^{6}-2x^{3}-2x^{2}+3x+c$

$2. \int (\frac{3}{x^{2}}-\frac{6}{x^{7}})dx=\int (3x^{-2}-6x^{-7})dx=\frac{3}{-2+1}x^{-2+1}-\frac{6}{-7+1}x^{-7+1} =-3x^{-1}+x^{-6}+c = -\frac{3}{x}+\frac{1}{x^{6}}+c$

$3. \int 6(\sqrt{x}+\sqrt[5]{x})dx=\int 6(x^{\frac{1}{2}}+x^{\frac{1}{5}})dx=6(\frac{1}{\frac{3}{2}}x^{\frac{3}{2}}+\frac{1}{\frac{6}{5}}x^{\frac{6}{5}}+c)=6(\frac{2}{3}x^{\frac{3}{2}}+\frac{5}{6}x^{\frac{6}{5}})+ c=4x^{\frac{3}{2}}+5x^{\frac{6}{5}}+c=4x\sqrt{x}+5x\sqrt[\sqrt[5]{x}]+c$

$4. \int \left ( \frac{1}{\sqrt{x}}-\frac{8}{\sqrt[3]{x}} \right )dx=\int \left ( x^{-\frac{1}{2}}-8x^{-\frac{1}{3}}\right) dx =\frac{1}{\frac{1}{2}}x^{\frac{1}{2}}-\frac{8}{\frac{2}{3}}x^{\frac{2}{3}} = 2\sqrt{x}-12\sqrt[3]{x^{2}}+c$

$5. \int \left \sqrt{2x} \right dx=\int \sqrt{2}\sqrt{x}dx=\sqrt{2}\int\ (x^{\frac{1}{2}})dx$
$= \sqrt{2}\frac{1}{\frac{3}{2}}x^{\frac{3}{2}}+c = \frac{2\sqrt{2}}{3}.x\sqrt{x}+c=\frac{2x\sqrt{2x}}{3}+c$

$6. \int \left ( \frac{dx}{\sqrt{2x}} \right )=\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{x}}=\frac{1}{\sqrt{2}}\int x^{-\frac{1}{2}}dx$

$= \frac{1}{\sqrt{2}}\frac{1}{\frac{1}{2}}x^{\frac{1}{2}}+c = \frac{2}{\sqrt{2}}\sqrt{x}+c=\sqrt{2x}+c$

$7. \int \left ( 3x-4 \right )^{2}dx=\int \left ( 9x^{2}-24x+16 \right )=x^{3}-12x^{2}+16x+c$

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MATEMATIKA INDONESIA

Selasa, 03 Agustus 2010

Integral Tak Tentu 2

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