Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$\displaystyle\int{ { e }^{ { \tan }^{ -1 } x } [ \frac{ 1+x+ { x }^{ 2 } }{ 1+ { x }^{ 2 } } ] }d x$$

Answer

$$(x*(e^(x/tan)[1*^(x/tan)[1+x+x^2))/(1+x^2)]d$$

Solution


Remove parentheses.
int(e)^(tan^-1*x)\(\frac{1+x+{x}^{2}}{1+{x}^{2}}\)d*x
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
int(e)^(1/tan*x)\(\frac{1+x+{x}^{2}}{1+{x}^{2}}\)d*x
Simplify  \(\frac{1}{tan}x\)  to  \(\frac{x}{tan}\).
int(e)^(x/tan)\(\frac{1+x+{x}^{2}}{1+{x}^{2}}\)d*x
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
[email protected]
Privacy Policy | Terms & Conditions
Contact Us