What is the integral of (x^2+8x+4)/(x^4+2x^3+x^2) ?

The integral of (x^2+8x+4)/(x^4+2x^3+x^2) is 3/(x+1)-4/x+C

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integral of (x^2+8x+4)/(x^4+2x^3+x^2)

\int \frac{x ^{2} + 8 x + 4}{x ^{4} + 2 x ^{3} + x ^{2}} d x

\frac{3}{x + 1} - \frac{4}{x} + C

Solution steps

\int \frac{x ^{2} + 8 x + 4}{x ^{4} + 2 x ^{3} + x ^{2}} d x

Expand

\frac{x ^{2} + 8 x + 4}{x ^{4} + 2 x ^{3} + x ^{2}} : \frac{x ^{2}}{x ^{4} + 2 x ^{3} + x ^{2}} + \frac{8 x}{x ^{4} + 2 x ^{3} + x ^{2}} + \frac{4}{x ^{4} + 2 x ^{3} + x ^{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{x ^{2}}{x ^{4} + 2 x ^{3} + x ^{2}} d x + \int \frac{8 x}{x ^{4} + 2 x ^{3} + x ^{2}} d x + \int \frac{4}{x ^{4} + 2 x ^{3} + x ^{2}} d x

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

\int \frac{8 x}{x ^{4} + 2 x ^{3} + x ^{2}} d x = 8 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

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

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

Simplify

- \frac{1}{x + 1} + 8 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) + 4 (- 2 ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ - \frac{1}{x + 1}) : \frac{3}{x + 1} - \frac{4}{x}

= \frac{3}{x + 1} - \frac{4}{x}

Add a constant to the solution

= \frac{3}{x + 1} - \frac{4}{x} + C

15 sin (- π x) tan (\frac{2}{3} π x)

f (x) = 3 x^{2} - 6 x + 5

\int_{0}^{\frac{π}{2}} 7 sin^{2} (2 x) d x

\int e^{7 x} (7) d x

x + y = 7, x + 5 = y^{2}

What is the integral of (x^2+8x+4)/(x^4+2x^3+x^2) ?
The integral of (x^2+8x+4)/(x^4+2x^3+x^2) is 3/(x+1)-4/x+C