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Jawaban Soal Teknik Integrasi Bab 7.1 Purcell

Carilah jawaban dari masalah!

1.\int (x-2)^5 dx =

Misal u=x-2 \Rightarrow \frac{du}{dx}=1\Rightarrow dx=du

\int (x-2)^5 dx = \int u^5 du = \frac{1}{6}u^6+C = \frac{1}{6}(x-2)^6+C

2.\int (3x)^{\frac{1}{2}} dx =\frac{1}{2}(3x)^{-\frac{1}{2}}+C

5. \int \frac{1}{x^2+4} dx = \int \frac{1}{x^2+2^2} dx=\frac{1}{2}tan^{-1}(\frac{x}{2})+C

8. \int \frac{2t^2}{2t^2+1}=

Catatan :


Proses :

\int \frac{2t^2}{2t^2+1}=\int 1- \frac{1}{2t^2+1}= t- \int \frac{1}{(\sqrt{2}t)^2+1}

Misalkan u=\sqrt{2}t \Rightarrow \frac{du}{dt}=\sqrt{2} \Rightarrow dt=\frac{du}{\sqrt{2}}

t- \int \frac{1}{(\sqrt{2}t)^2+1}=t- \frac{1}{\sqrt{2}}tan^{-1}\sqrt{2}t+C

11. \int \frac{tan z}{cos^2z}dz =

Misalkan u= tan z \Rightarrow \frac{du}{dz}=\frac{1}{cos^2z}

13. \int \frac {sin \sqrt {t}}{\sqrt{t}} dt=

Misalka u= \sqrt{t}

14. \int \frac{2x}{\sqrt{1-x^4}}dx=

\int \frac{2x}{\sqrt{1-x^4}}dx=\int \frac{2x}{\sqrt{1-(x^2)^2}}dx

Misalkan u=x^2

15. \int \frac{cos x}{1+sin^2 x}dx

Misalkan u=sin x

17 \int \frac{3x^2+2x}{x+1}dx=

Lakukan pembagian biasa sehingga diperoleh hasil 3x dengan sisa \frac {-x}{x+1}

Rubah \frac {-x}{x+1} menjadi \frac{1}{x+1}-1

19. \int \frac{sin (ln 4x^2)}{x}dx

Misal u=ln4x^2

21. \int \frac{6e^x}{\sqrt{1-e^{2x}}}dx=

\int \frac{6e^x}{\sqrt{1-e^{2x}}}dx=\int \frac{6e^x}{\sqrt{1-(e^{x})^2}}dx

27. \int \frac{sin x -cos x}{sin x}dx =

\int \frac{sin x -cos x}{sin x}dx =\int \frac{sin x }{sin x}- \frac{cos x}{sin x}dx

28. \frac{sin (4t-1)}{1-sin^2(4t-1)}dt =

ganti 1-sin^2(4t-1) menjadi cos^2(4t-1)

32. \int \frac{(6t-1)sin \sqrt{3t^2-t-1}}{\sqrt{3t^2-t-1}}

Misalkan u=3t^2-t-1

33. \int \frac{t^2cos (t^3-2)}{sin^2(t^3-2)}dt

Misalkan u=sin( t^3-2)

34. \int \frac{1+cos 2x }{sin^2 2x}=

\int \frac{1+cos 2x }{sin^2 2x}=\int \frac{1}{sin^2 2x}+ \int \frac{cos 2x}{sin^2 2x}

37. \int \frac{e^{tan-1 2t}}{1+4t^2}dt=

Misal u=tan^-1 2t

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