What is the integral of sqrt(1+x^{-2)} ?

The integral of sqrt(1+x^{-2)} is -1/2 ln|sqrt(x^2+1)+1|+1/2 ln|sqrt(x^2+1)-1|+sqrt(x^2+1)+C

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integral of sqrt(1+x^{-2)}

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

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

Solution steps

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

Simplify

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

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

Simplify

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

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

x^{2} = (x),

assuming

x \geq 0

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

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} - 1} d u

= \int \frac{u ^{2}}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

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}{- u ^{2} + 1} d u - \int 1 d u)

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u)

Substitute back

u = x^{2} + 1

= - (\frac{ln x ^{2} + 1 + 1}{2} - \frac{ln x ^{2} + 1 - 1}{2} - x^{2} + 1)

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{x y - 1}{x} d x

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

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

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