Indefinite Integral

We have covered the introduction to derivatives. Another topic would be to find a function if we know it’s derivative.

This very interesting and we won’t spend much time explaining what the antiderivative is in that case.

A function $F$ is an antiderivative of the function $f$ if $F'(x)=f(x)$ for all $x$ in the domain of $f$ .

The only problem is when we calculated the derivative of a constant $C$ , we found $0$ .

Let’s check the following fact:

If we have $F'(x)=3x^{2}$

We know that $(x^{3})'=3x^{2}$

But so is: $(x^{3}+1000)'$ .

And we can add any constant, the derivative will be the same.

We get the theorem:

Theorem: General form of antiderivatives

Let $F$ be an antiderivative of $f$ over an interval $I$ . Then,

1. For each constant $C$ , the function $F(x)+C$ is also an antiderivative of $f$ over $I$ .

2. If $G$ is an antiderivative of $f$ over $I$ , there is a constant $C$ for which $G(x)=F(x)+C$ over $I$ .

Meaning, the general form of the antiderivative of $f$ over $I$ is $F(x)+C$ .

EXAMPLE:

Find the antiderative of $6x$

We can see that it is $3x^{2}+C$

Indefinite Integrals

This is the notation used for the antiderivatives.

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

$f(x)$ is called the $integrand$ and the variable $x$ is the $variable \; of \; integration$

Fundamental formulas

$1. {\displaystyle \int \frac{d}{dx}[f(x)] \,dx}=f(x)+C$

$2. {\displaystyle \int [f(x)+g(x)] \,dx}=\int f(x) dx+ \int g(x) dx$

$3. {\displaystyle \int af(x)dx}=a \int f(x) \,dx$ with $a$ a constant

$4. {\displaystyle \int \frac{dx}{x}}=\ln |x|+C$

$5. {\displaystyle \int x^{n} \,dx}=\frac{x^{n+1}}{n+1}+C$

$6. {\displaystyle \int e^{x} \,dx}=e^{x}+C$

$7. {\displaystyle \int a^{x} \,dx}=\frac{a^{x}}{\ln a}+C$ with $a>0$ , $a \neq 1$

$8. {\displaystyle \int \sin x \,dx}=-\cos x +C$

$9. {\displaystyle \int \cos x \,dx}=\sin x +C$

$10. {\displaystyle \int \tan x\, dx}=\ln |\sec x|+C$

$11. {\displaystyle \int \cot x\, dx}=\ln |\sin x|+C$

$12. {\displaystyle \int \sec x\, dx}=\ln |\sec x +\tan x|+C$

$13. {\displaystyle \int \csc x\, dx}=\ln |\csc x -\cot x|+C$

$14. {\displaystyle \int \sec^{2} x\, dx}=\tan x + C$

$15. {\displaystyle \int \frac{dx}{1+x^{2}}}= \tan^{-1} x +C$

$16. {\displaystyle \int \frac{dx}{a^{2}+x^{2}}} = \frac{1}{a}\tan^{-1} \frac{x}{a} +C$

$17. {\displaystyle \int \frac{dx}{x^{2}-a^{2}}}= \frac{1}{2a} \ln |\frac{x-a}{x+a}|+C$

$18. {\displaystyle \int \frac{dx}{a^{2}-x^{2}}}= \frac{1}{2a} \ln |\frac{a+x}{a-x}|+C$

$19. {\displaystyle \int \frac{dx}{\sqrt{a^{2}-x^{2}}}}= \sin ^{-1} \frac{x}{a}+C$

$20. {\displaystyle \int \frac{dx}{\sqrt{x^{2} \pm a^{2}}}=\ln |x+\sqrt{x^{2} \pm a^{2}} |}+C$

Problem 1:

Solve: $\int \frac{x+3}{x^{2}-2x-5}\, dx$

We know that $\frac{d}{dx} (x^{2}-2x-5)=2x-2$

From $x+3$ we can do some inspection

$x+3=\frac{1}{2}(2x+6)=\frac{1}{2}(2x+8-2)=\frac{1}{2}(2x-2)+4$

We get

(1) $\begin{equation*} \begin{split} \displaystyle \int \frac{x+3}{x^{2}-2x-5}\, dx}&={\displaystyle \int \frac{\frac{1}{2}(2x-2)+4}{x^{2}-2x-5}\, dx}\\ &=\frac{1}{2}{\displaystyle \int \frac{2x-2}{x^{2}-2x-5}\,dx}+4{\displaystyle \int \frac{dx}{x^{2}-2x-5}}\\ &= \frac{1}{2}{\displaystyle \int \frac{2x-2}{x^{2}-2x-5}\,dx}+4{\displaystyle \int \frac{dx}{(x-1)^{2}-(\sqrt{6})^{2}}}\\ &= \frac{1}{2} \ln |x^{2}-2x-5|+4\cdot \frac{1}{2 \sqrt{6}} \ln \left |\frac{(x-1)-\sqrt{6}}{(x-1)+\sqrt{6}} \right | +C \end{split} \end{equation*}$

Finally:

${\displaystyle \int \frac{x+3}{x^{2}-2x-5}\, dx}=\frac{1}{2} \ln |x^{2}-2x-5|+2\cdot \frac{1}{ \sqrt{6}} \ln \left |\frac{(x-1)-\sqrt{6}}{(x-1)+\sqrt{6}} \right | +C$

Problem 2:

Solve: $\int \frac{3x+2}{\sqrt{x^{2}+5x+12}}\, dx$

We know that $\frac{d}{dx} (x^{2}+5x+12)=2x+5$

From $3x+2$ we can do some inspection

$3x+2=\frac{3}{2}(2x+\frac{4}{3})=\frac{3}{2}(2x+5-\frac{11}{3})=\frac{3}{2}(2x+5)-\frac{11}{2}$

We get

(2) $\begin{equation*} \begin{split} \displaystyle \int \frac{3x+2}{\sqrt{x^{2}+5x+12}}\, dx}&= {\displaystyle \int \frac{\frac{3}{2}(2x+5)-\frac{11}{2}}{\sqrt{x^{2}+5x+12}}\, dx}\\ &=\frac{3}{2}{\displaystyle \int \frac{(2x+5)}{\sqrt{x^{2}+5x+12}}\, dx}-\frac{11}{2} {\displaystyle \int \frac{dx}{\sqrt{x^{2}+5x+12}}}\\ &=\frac{3}{2}{\displaystyle \int \frac{(2x+5)}{\sqrt{x^{2}+5x+12}}\, dx}-\frac{11}{2} {\displaystyle \int \frac{dx}{\sqrt{(x+\frac{5}{2})^{2}+\frac{23}{4}}}} \\ &=\frac{3}{2} \ln |x^{2}+5x+12|-\frac{11}{2} \ln | x+ \frac{5}{2}+\sqrt{x^{2}+5x+12}| +C \end{split} \end{equation*}$