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

The integral of 1/((x^2+1)^3) is 3/8 (arctan(x)+x/(1+x^2))+x/(4(x^2+1)^2)+C

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

integral

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

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

Solution steps

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

Apply Integral Reduction

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

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

= \frac{3}{4} \cdot \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{2 \cdot 2 ( x ^{2} + 1 ) ^{2}}

Simplify

\frac{3}{4} \cdot \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{2 \cdot 2 ( x ^{2} + 1 ) ^{2}} : \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}}

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

Add a constant to the solution

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

y (4 - y)^{\frac{1}{2}}

(\frac{e}{2})^{x}

\frac{1}{( x - a )}

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

\frac{8 x}{( x ^{4} + 1 )}

What is the integral of 1/((x^2+1)^3) ?
The integral of 1/((x^2+1)^3) is 3/8 (arctan(x)+x/(1+x^2))+x/(4(x^2+1)^2)+C