What is the integral of (e^x+x^3-1/(x^4)+1/(1+x^2)) ?

The integral of (e^x+x^3-1/(x^4)+1/(1+x^2)) is e^x+(x^4)/4+1/(3x^3)+arctan(x)+C

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integral of (e^x+x^3-1/(x^4)+1/(1+x^2))

\int (e^{x} + x^{3} - \frac{1}{x ^{4}} + \frac{1}{1 + x ^{2}}) d x

e^{x} + \frac{x ^{4}}{4} + \frac{1}{3 x ^{3}} + arctan (x) + C

Solution steps

\int e^{x} + x^{3} - \frac{1}{x ^{4}} + \frac{1}{1 + x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{x} d x + \int x^{3} d x - \int \frac{1}{x ^{4}} d x + \int \frac{1}{1 + x ^{2}} d x

\int e^{x} d x = e^{x}

\int x^{3} d x = \frac{x ^{4}}{4}

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

= e^{x} + \frac{x ^{4}}{4} - (- \frac{1}{3 x ^{3}}) + arctan (x)

Simplify

= e^{x} + \frac{x ^{4}}{4} + \frac{1}{3 x ^{3}} + arctan (x)

Add a constant to the solution

= e^{x} + \frac{x ^{4}}{4} + \frac{1}{3 x ^{3}} + arctan (x) + C

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What is the integral of (e^x+x^3-1/(x^4)+1/(1+x^2)) ?
The integral of (e^x+x^3-1/(x^4)+1/(1+x^2)) is e^x+(x^4)/4+1/(3x^3)+arctan(x)+C