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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cot^{2} (u) sec (u) d u

= \int cot^{2} (u) sec (u) d u

Simplify

cot^{2} (u) sec (u) : cot (u) csc (u)

= \int cot (u) csc (u) d u

Apply u-substitution

: \int - 1 d v

= \int - 1 d v

Integral of a constant:

\int a d x = a x

= (- 1) v

Substitute back

= (- 1) csc (arctan (x))

Simplify

(- 1) csc (arctan (x)) : - \frac{1 + x ^{2}}{x}

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

Add a constant to the solution

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

\int \frac{8 x}{2 x ^{2} + 1} d x

y = (x + 1)^{3}

\int \frac{x}{x ^{2} - 2} d x

\int \frac{1}{16 + 25 x ^{2}} d x

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

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