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

The integral of x/((x^2+1)(x^2+4x+4)) is 1/25 (3/2 ln|x^2+1|+4arctan(x))-3/25 ln|x+2|+2/(5(x+2))+C

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

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

\frac{1}{25} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{25} ln ∣ x + 2 ∣ + \frac{2}{5 ( x + 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 4 x + 4 )} : \frac{3 x + 4}{25 ( x ^{2} + 1 )} - \frac{3}{25 ( x + 2 )} - \frac{2}{5 ( x + 2 ) ^{2}}

= \int \frac{3 x + 4}{25 ( x ^{2} + 1 )} - \frac{3}{25 ( x + 2 )} - \frac{2}{5 ( x + 2 ) ^{2}} d x

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 + 4}{25 ( x ^{2} + 1 )} d x - \int \frac{3}{25 ( x + 2 )} d x - \int \frac{2}{5 ( x + 2 ) ^{2}} d x

\int \frac{3 x + 4}{25 ( x ^{2} + 1 )} d x = \frac{1}{25} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x))

\int \frac{3}{25 ( x + 2 )} d x = \frac{3}{25} ln ∣ x + 2 ∣

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

= \frac{1}{25} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{25} ln ∣ x + 2 ∣ - (- \frac{2}{5 ( x + 2 )})

Simplify

= \frac{1}{25} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{25} ln ∣ x + 2 ∣ + \frac{2}{5 ( x + 2 )}

Add a constant to the solution

= \frac{1}{25} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{25} ln ∣ x + 2 ∣ + \frac{2}{5 ( x + 2 )} + C

x \to 0 lim (8 x csc (5 x))

11 (t + 1) \frac{d y}{d t} - 8 y = 24 t

y^{^{'}} + e^{x} y = e^{x}

\frac{d}{d t} (t - \frac{1}{t})

\frac{d y}{d x} = e^{\frac{y}{x}} + \frac{y}{x}

What is the integral of x/((x^2+1)(x^2+4x+4)) ?
The integral of x/((x^2+1)(x^2+4x+4)) is 1/25 (3/2 ln|x^2+1|+4arctan(x))-3/25 ln|x+2|+2/(5(x+2))+C