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Wednesday, August 8, 2012

INTEGRAL TAK TENTU

RUMUS-RUMUS :

\[\int x^n\:dx =\frac{x^{n\,+\,1}}{n+1}+C\]berlaku untuk semua nilai n kecuali n = - 1

\[\int \frac{1}{x}\:dx =\ln\:x+C\]

\[\int e^x\:dx= e^x+C\]

\[\int \sin\:x\:dx = -\cos\:x +C\]

\[\int \cos\:x\:dx = \;\sin\:x +C\]

\[\int \tan\:x\:dx= -\ln \cos x +C\]

\[\int \sec^2\:x\:dx= \tan\:x +C\]

\[\int \frac{1}{a^2 + x^2}\:dx = \frac{1}{a} \tan^{-1}\frac{x}{a} +C\]

\[\int \frac{1}{a^2 - x^2}\:dx= \frac{1}{2a} \ln \left(\frac{a + x}{a - x}\right)+C= \frac{1}{a} \tanh^{-1}\frac{x}{a} +C\]

\[\int \frac{1}{x^2 - a^2}\:dx= \frac{1}{2\,a} \ln \left(\frac{x-a}{x+a} \right)+C= -\frac{1}{a} \coth^{-1}\frac{x}{a}+C\]

\[\int \frac{1}{\sqrt{(a^2\:-\:x^2)}}\:dx = \sin^{-1} \frac{x}{a}+C\]

\[\int \frac{1}{\sqrt{(a^2\:+\:x^2)}}\:dx= \ln\left(x + \sqrt{(x^2\:+\:a^2)} \right) = \sinh^{-1}\frac{x}{a}+C\]

\[\int \frac{1}{\sqrt{(x^2\:-\:a^2)}}\:dx= \ln\left(x\:+\:\sqrt{(x^2\:-\:a^2}) \right)+C\]

2 comments:

  1. mas tentukan hasil integral berikut :
    \int \left ( 2x-3 \right )^{9}dx
    terimakasih sebelumnya

    ReplyDelete
    Replies
    1. soal ini masuk kategori integral substitusi:
      \int \left ( 2x-3 \right )^{9}dx=\frac{1}{20}\left ( 2x-3 \right )^{10}+C

      Delete

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