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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x}{( x ^{2} + 1 ) ^{2}} - \int - \frac{6 x ^{2}}{( x ^{2} + 1 ) ^{3}} d x

= \frac{x}{( x ^{2} + 1 ) ^{2}} - \int - \frac{6 x ^{2}}{( x ^{2} + 1 ) ^{3}} d x

\int - \frac{6 x ^{2}}{( x ^{2} + 1 ) ^{3}} d x = - \frac{3}{4} (arctan (x) - \frac{1}{4} sin (4 arctan (x)))

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{2} \frac{x ^{2} - 2}{x + 1} d x

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 0

\frac{d}{d x} (cos^{7} (x))

y^{^{'}} = \frac{x ^{2} y + y ^{3}}{x ^{3}}

\frac{\partial}{\partial y} (- sin (y))

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