What is the integral of 1/((1+x^2)^{0.5)} ?

The integral of 1/((1+x^2)^{0.5)} is ln|x+sqrt(x^2+1)|+C

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integral of 1/((1+x^2)^{0.5)}

integral

\frac{1}{( 1 + x ^{2} ) ^{0.5}}

ln x + x^{2} + 1 + C

Solution steps

\int \frac{1}{( 1 + x ^{2} ) ^{0.5}} d x

Simplify

x^{2} : xx

= \int \frac{1}{( 1 + xx ) ^{0.5}} d x

xx = x^{2}

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

Apply Integration By Parts

: \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

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

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

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

Simplify

= ln x + x^{2} + 1

Add a constant to the solution

= ln x + x^{2} + 1 + C

\frac{- 1}{y + 1}

x^{2} e^{2} x

\frac{3}{x ^{4}}

\frac{e ^{x}}{x ^{4}}

\frac{tan ( x )}{1 + cos ^{4} ( x )}

What is the integral of 1/((1+x^2)^{0.5)} ?
The integral of 1/((1+x^2)^{0.5)} is ln|x+sqrt(x^2+1)|+C