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

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

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integral of (4x^3-3x)/((x^2+1)^2)

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int_{2}^{3} x^{2} e^{x^{3} - 8} d x

\frac{x ^{2}}{x + 7}

49 y^{^{''}} - 16 y = 0

\frac{3 ^{2}}{( s + 3 ) ^{2}}

\int 1 + cos (t) d t

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