Indefinite Integrals

1. Basic rules of integration

1) If F'(x)=f(x), then

\int f(x) dx=F(x)+C

Where C is an arbitrary constant.

2) \int{Af(x)dx}=A\int{f(x)dx} , where A  is a constant quantity.

3) \int \left[ f_1(x) \pm f_2(x) dx \right] =\int f_1(x) dx\pm \int f_2(x) dx

4) If  \int f(x) dx=F(x)+C and u=\phi (x), then

\int f(u) dx=F(u)+C

In particular,

\displaystyle \int f(ax+b) dx=\frac{1}{a}F(ax+b)+C

2. Table of standard integrals

I. \displaystyle \int x^n dx = \frac{x^{n+1}}{n+1}; n\ne -1
II. \displaystyle \int \frac{dx}{x} =\ln |x|+C
III. \displaystyle \int \frac{dx}{x^2+a^2} dx =\frac{1}{a}\arctan \frac{x}{a}+C; (a \ne 0)
IV. \displaystyle \int \frac{dx}{x^2-a^2} =\frac{1}{2a} \ln \left|\frac{x-a}{x+a}\right|+C; (a \ne 0)
\displaystyle \int \frac{dx}{a^2-x^2} =\frac{1}{2a} \ln \left|\frac{a+x}{a-x}\right|+C ; (a \ne 0)
V. \displaystyle \int \frac{dx}{\sqrt{x^2+a}} = \ln \left|x+\sqrt{x^2+a}\right|+C ; (a \ne 0)
VI. \displaystyle \int \frac{dx}{\sqrt{a^2-x^2}} = \arcsin \frac{x}{a}+C=-\arccos \frac{x}{a}+C; a>0
VII. \displaystyle \int a^xdx = \frac{a^x}{\ln a}+C, a>0; \int e^xdx=e^x+C
VIII. \int \sin x dx =-\cos x +C
IX. \int \cos x dx =\sin x +C
X. \displaystyle \int \frac{dx}{\cos^2 x}=\tan x +C
XI. \displaystyle \int \frac{dx}{\sin^2 x}=-\cot x +C
XII. \displaystyle \int \frac{dx}{\sin x}=\ln \left| \tan \frac{x}{2}\right|+C=\ln \left| \frac{1}{\sin x}-\cot x \right|+C
XIII \displaystyle \int \frac{dx}{\cos x}=\ln \left| \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right|+C=\ln \left| \frac{1}{\cos x}+\tan x \right|+C
XIV. \displaystyle \int \sinh x dx =\cosh x +C; \sinh x =\frac{1}{2}\left(e^x-e^{-x}\right); \cosh x =\frac{1}{2}\left(e^x+e^{-x}\right)
XV. \int \cosh x dx =\sinh x +C
XVI. \displaystyle \int \frac{dx}{\cosh^2 x}=\tanh x +C; \tanh x =\frac{\sinh x}{\cosh x}
XVII. \displaystyle \int \frac{dx}{\sinh^2 x}=-\coth x +C; \coth x =\frac{\cosh x}{\sinh x}

Related Posts

Comments

  1. This is such a great resource that you are providing and you give it away for free. I enjoy seeing websites that understand the value of providing a prime resource for free. I truly loved reading your post. Thanks!

Speak Your Mind

*