What is the integral of ((3x^3+4x))/((x^2+5)^2) ?

The integral of ((3x^3+4x))/((x^2+5)^2) is 3/2 (ln|x^2+5|+5/(x^2+5))-2/(x^2+5)+C

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integral of ((3x^3+4x))/((x^2+5)^2)

\int \frac{( 3 x ^{3} + 4 x )}{( x ^{2} + 5 ) ^{2}} d x

\frac{3}{2} (ln x^{2} + 5 + \frac{5}{x ^{2} + 5}) - \frac{2}{x ^{2} + 5} + C

Solution steps

\int \frac{3 x ^{3} + 4 x}{( x ^{2} + 5 ) ^{2}} d x

Expand

\frac{3 x ^{3} + 4 x}{( x ^{2} + 5 ) ^{2}} : \frac{3 x ^{3}}{( x ^{2} + 5 ) ^{2}} + \frac{4 x}{( x ^{2} + 5 ) ^{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{3 x ^{3}}{( x ^{2} + 5 ) ^{2}} d x + \int \frac{4 x}{( x ^{2} + 5 ) ^{2}} d x

\int \frac{3 x ^{3}}{( x ^{2} + 5 ) ^{2}} d x = \frac{3}{2} (ln x^{2} + 5 + \frac{5}{x ^{2} + 5})

\int \frac{4 x}{( x ^{2} + 5 ) ^{2}} d x = - \frac{2}{x ^{2} + 5}

= \frac{3}{2} (ln x^{2} + 5 + \frac{5}{x ^{2} + 5}) - \frac{2}{x ^{2} + 5}

Add a constant to the solution

= \frac{3}{2} (ln x^{2} + 5 + \frac{5}{x ^{2} + 5}) - \frac{2}{x ^{2} + 5} + C

((x^{2} + 1)^{2})^{^{'}}

y^{^{''}} + 4 y^{^{'}} + 4 y = sec (t) tan (t)

\int_{0}^{1} (x^{2} - x) d x

3000 (14 - x) (x - 5)

f (x) = 6 x^{2} + 8 x

What is the integral of ((3x^3+4x))/((x^2+5)^2) ?
The integral of ((3x^3+4x))/((x^2+5)^2) is 3/2 (ln|x^2+5|+5/(x^2+5))-2/(x^2+5)+C