What is the integral of 1/(1+\frac{1){x^2}} ?

The integral of 1/(1+\frac{1){x^2}} is -arctan(x)+x+C

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integral of 1/(1+\frac{1){x^2}}

\int \frac{1}{1 + \frac{1}{x ^{2}}} d x

- arctan (x) + x + C

Solution steps

\int \frac{1}{1 + \frac{1}{x ^{2}}} d x

Simplify

\frac{1}{1 + \frac{1}{x ^{2}}} : \frac{x ^{2}}{x ^{2} + 1}

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

\frac{x ^{2}}{x ^{2} + 1} = - \frac{1}{x ^{2} + 1} + 1

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

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

\int 1 d x = x

= - arctan (x) + x

Add a constant to the solution

= - arctan (x) + x + C

\int \frac{sec ^{2} ( 5 x )}{1 - tan ( 5 x )} d x

\int (1 + tan (x)) sec^{2} (x) d x

\int sin (46 x) sec^{2} (cos (46 x)) d x

\int \frac{sin ^{2} ( x )}{sin ( x )} d x

What is the integral of 1/(1+\frac{1){x^2}} ?
The integral of 1/(1+\frac{1){x^2}} is -arctan(x)+x+C