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

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

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

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

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

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

\int \frac{x ^{2}}{x ^{3} + 2} d x

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

7

y^{^{'}} + a y + b = 0

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