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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: - \frac{1 + x}{x} - \int - \frac{1}{2 x 1 + x} d x

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (3 9 x + 1)

- 6 \cdot x^{2} \cdot y + \frac{d y}{d x} = 5 x^{2}

\frac{\partial}{\partial x} (3 x^{4} + 5 y^{4} + 9 x y)

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{x}

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

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