What is the integral of sqrt((1+x^2)^3) ?

The integral of sqrt((1+x^2)^3) is 1/4 x(x^2+1)^{3/2}+3/8 ln|x+sqrt(x^2+1)|+3/8 xsqrt(x^2+1)+C

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integral of sqrt((1+x^2)^3)

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

\frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} ln x + x^{2} + 1 + \frac{3}{8} x x^{2} + 1 + C

Solution steps

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

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

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

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

Apply Integration By Parts

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

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

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

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

Simplify

x (1 + x^{2})^{\frac{3}{2}} - 3 (- \frac{1}{4} (\frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2}) + \frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}}) : \frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} ln x + x^{2} + 1 + \frac{3}{8} x x^{2} + 1

= \frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} ln x + x^{2} + 1 + \frac{3}{8} x x^{2} + 1

Add a constant to the solution

= \frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} ln x + x^{2} + 1 + \frac{3}{8} x x^{2} + 1 + C

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What is the integral of sqrt((1+x^2)^3) ?
The integral of sqrt((1+x^2)^3) is 1/4 x(x^2+1)^{3/2}+3/8 ln|x+sqrt(x^2+1)|+3/8 xsqrt(x^2+1)+C