What is the integral of 1/((1+x^2)sqrt(1+x^2)) ?

The integral of 1/((1+x^2)sqrt(1+x^2)) is x/(sqrt(1+x^2))+C

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integral of 1/((1+x^2)sqrt(1+x^2))

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

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

Solution steps

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

Simplify

(1 + x^{2}) 1 + x^{2} : (1 + x^{2})^{\frac{3}{2}}

= \int \frac{1}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= sin (u)

Substitute back

u = arctan (x)

= sin (arctan (x))

Use the following identity:

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

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

Add a constant to the solution

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

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What is the integral of 1/((1+x^2)sqrt(1+x^2)) ?
The integral of 1/((1+x^2)sqrt(1+x^2)) is x/(sqrt(1+x^2))+C