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

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

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

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

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

Solution steps

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

Simplify

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

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{3} - 2 x + 2, (2, 6)

4 t \frac{d y}{d t} - 3 y = t

\frac{d}{d x} (- 2 x sin (2 x) + cos (2 x))

\frac{( s + 1 )}{s ( s ^{2} + s + 1 )}

\frac{d y}{d x} + 1 = 2 y

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