What is the integral of (34x^2)/((2x+1)(x^2+4)) ?

The integral of (34x^2)/((2x+1)(x^2+4)) is ln|2x+1|+8ln|(x^2)/4+1|-4arctan(x/2)+C

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integral of (34x^2)/((2x+1)(x^2+4))

\int \frac{34 x ^{2}}{( 2 x + 1 ) ( x ^{2} + 4 )} d x

ln ∣ 2 x + 1 ∣ + 8 ln \frac{x ^{2}}{4} + 1 - 4 arctan (\frac{x}{2}) + C

Solution steps

\int \frac{34 x ^{2}}{( 2 x + 1 ) ( x ^{2} + 4 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 34 \cdot \int \frac{x ^{2}}{( 2 x + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{x ^{2}}{( 2 x + 1 ) ( x ^{2} + 4 )} : \frac{1}{17 ( 2 x + 1 )} + \frac{8 x - 4}{17 ( x ^{2} + 4 )}

= 34 \cdot \int \frac{1}{17 ( 2 x + 1 )} + \frac{8 x - 4}{17 ( x ^{2} + 4 )} 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

= 34 (\int \frac{1}{17 ( 2 x + 1 )} d x + \int \frac{8 x - 4}{17 ( x ^{2} + 4 )} d x)

\int \frac{1}{17 ( 2 x + 1 )} d x = \frac{1}{34} ln ∣ 2 x + 1 ∣

\int \frac{8 x - 4}{17 ( x ^{2} + 4 )} d x = \frac{2}{17} (2 ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2}))

= 34 (\frac{1}{34} ln ∣ 2 x + 1 ∣ + \frac{2}{17} (2 ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})))

Simplify

34 (\frac{1}{34} ln ∣ 2 x + 1 ∣ + \frac{2}{17} (2 ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2}))) : ln ∣ 2 x + 1 ∣ + 8 ln \frac{x ^{2}}{4} + 1 - 4 arctan (\frac{x}{2})

= ln ∣ 2 x + 1 ∣ + 8 ln \frac{x ^{2}}{4} + 1 - 4 arctan (\frac{x}{2})

Add a constant to the solution

= ln ∣ 2 x + 1 ∣ + 8 ln \frac{x ^{2}}{4} + 1 - 4 arctan (\frac{x}{2}) + C

y^{^{'}} = - 3 x y^{3}

\int (6 x y) d x

e^{3 x}, 2

\frac{d}{d x} (x^{p (x)})

w

What is the integral of (34x^2)/((2x+1)(x^2+4)) ?
The integral of (34x^2)/((2x+1)(x^2+4)) is ln|2x+1|+8ln|(x^2)/4+1|-4arctan(x/2)+C