 gaiaecologicaq2

2022-07-09

Find The Antiderivative of more complex expressions.
Need help finding the antiderivative of this one in particular $\frac{dx}{3x²+1}$.
Would you kindly also give some tips on how to find the antiderivative of any expressions. I am having problems switching from differential calculus to integral calculus as finding the antiderivative is not as straight forward as finding the differential of an expression. pampatsha

Expert

Step 1
$I=\int \frac{\mathrm{d}x}{{a}^{2}+\left(bx{\right)}^{2}}$
Let $z=\frac{bx}{a}$ so that $\mathrm{d}z=\frac{b}{a}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$. Thus
$\begin{array}{rl}I& =\frac{a}{b}\int \frac{\mathrm{d}z}{{a}^{2}+{a}^{2}{z}^{2}}\\ & =\frac{1}{ab}\int \frac{\mathrm{d}z}{1+{z}^{2}}\end{array}$
Step 2
The integral is now easily seen to be $\mathrm{arctan}z+C$. Thus
$\int \frac{\mathrm{d}x}{{a}^{2}+\left(bx{\right)}^{2}}=\frac{1}{ab}\mathrm{arctan}\left(\frac{bx}{a}\right)+C$
Now set $a=1$ and $b=\sqrt{3}$. Holetaug

Expert

Step 1
You need to know the formula
$\int \frac{1}{{x}^{2}+{a}^{2}}dx=\frac{1}{a}\mathrm{arctan}\left(\frac{x}{a}\right)+c$
$\begin{array}{rl}\int \frac{1}{3{x}^{2}+1}dx& =\frac{1}{3}\int \frac{1}{{x}^{2}+\frac{1}{3}}dx=\frac{1}{3}\int \frac{1}{{x}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}dx\\ & =\frac{1}{3}×\frac{1}{1/\sqrt{3}}\mathrm{arctan}\left(\frac{x}{1/\sqrt{3}}\right)+c\\ & =\frac{1}{\sqrt{3}}\mathrm{arctan}\sqrt{3}x+c\end{array}$
Step 2
ANOTHER METHOD: By Substitution
Let $x=\frac{1}{\sqrt{3}}\mathrm{tan}\theta$. So . Also $3{x}^{2}+1=3\left(\frac{1}{3}{\mathrm{tan}}^{2}\theta \right)+1={\mathrm{tan}}^{2}\theta +1={\mathrm{sec}}^{2}\theta$.
Now since $x=\frac{1}{\sqrt{3}}\mathrm{tan}\theta$, so $\theta =\mathrm{arctan}\sqrt{3}x$. Putting this in one we have the required answer as $\int \frac{1}{3{x}^{2}+1}dx=\frac{1}{\sqrt{3}}\mathrm{arctan}\sqrt{3}x+c$.