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

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

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

integral

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

\frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - 2 ln x + 1 + x^{2} + \frac{2 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x ( x ^{3} + 2 x )}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x = \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + 2 (ln x + 1 + x^{2} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}})

= \frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + 2 (ln x + 1 + x^{2} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}})

Simplify

\frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + 2 (ln x + 1 + x^{2} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}) : \frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - 2 ln x + 1 + x^{2} + \frac{2 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - 2 ln x + 1 + x^{2} + \frac{2 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x ^{3}}{( x ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 + \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} - 2 ln x + 1 + x^{2} + \frac{2 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

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

1 + tan (x)

3 x^{2} + 8 x + 6

x^{2} + 2

e^{x} \cdot sin^{2} (x)

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