What is the integral of 1/(sqrt(ax+b)) ?

The integral of 1/(sqrt(ax+b)) is x/(sqrt(ax+b))+(sqrt(ax+b))/a+b/(asqrt(ax+b))+C

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integral of 1/(sqrt(ax+b))

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

\frac{x}{a x + b} + \frac{a x + b}{a} + \frac{b}{a a x + b} + C

Solution steps

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

Apply Integration By Parts

: \frac{x}{a x + b} - \int - \frac{a x}{2 ( a x + b ) ^{\frac{3}{2}}} d x

= \frac{x}{a x + b} - \int - \frac{a x}{2 ( a x + b ) ^{\frac{3}{2}}} d x

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

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

Simplify

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

= \frac{x}{a x + b} + \frac{a x + b}{a} + \frac{b}{a a x + b}

Add a constant to the solution

= \frac{x}{a x + b} + \frac{a x + b}{a} + \frac{b}{a a x + b} + C

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = \frac{e ^{x}}{x ^{2} + 1}

y = ln (x^{2} + y^{2})

\frac{\partial}{\partial y} (e^{x^{2} - y^{2}})

\frac{d}{d x} (1.5 (\frac{1}{3} + \frac{x}{2} - \frac{x ^{3}}{6}))

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

What is the integral of 1/(sqrt(ax+b)) ?
The integral of 1/(sqrt(ax+b)) is x/(sqrt(ax+b))+(sqrt(ax+b))/a+b/(asqrt(ax+b))+C