What is the integral of (x^2+1)/(x^4+10x^3+21x^2) ?

The integral of (x^2+1)/(x^4+10x^3+21x^2) is

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integral of (x^2+1)/(x^4+10x^3+21x^2)

\int \frac{x ^{2} + 1}{x ^{4} + 10 x ^{3} + 21 x ^{2}} d x

\frac{x}{x ^{2} + 10 x + 21} + 8 (\frac{1}{32} ln ∣ 2 x + 6 ∣ + \frac{3}{16 ( 2 x + 6 )} - \frac{1}{32} ln ∣ 2 x + 14 ∣ - \frac{7}{16 ( 2 x + 14 )}) - \frac{10}{441} ln ∣ x ∣ - \frac{1}{21 x} + \frac{1}{36} ln ∣ x + 3 ∣ - \frac{1}{196} ln ∣ x + 7 ∣ + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{4} + 10 x ^{3} + 21 x ^{2}} d x

Expand

\frac{x ^{2} + 1}{x ^{4} + 10 x ^{3} + 21 x ^{2}} : \frac{x ^{2}}{x ^{4} + 10 x ^{3} + 21 x ^{2}} + \frac{1}{x ^{4} + 10 x ^{3} + 21 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} + 10 x ^{3} + 21 x ^{2}} d x + \int \frac{1}{x ^{4} + 10 x ^{3} + 21 x ^{2}} d x

\int \frac{x ^{2}}{x ^{4} + 10 x ^{3} + 21 x ^{2}} d x = \frac{x}{x ^{2} + 10 x + 21} + 8 (\frac{1}{32} ln ∣ 2 x + 6 ∣ + \frac{3}{16 ( 2 x + 6 )} - \frac{1}{32} ln ∣ 2 x + 14 ∣ - \frac{7}{16 ( 2 x + 14 )})

\int \frac{1}{x ^{4} + 10 x ^{3} + 21 x ^{2}} d x = - \frac{10}{441} ln ∣ x ∣ - \frac{1}{21 x} + \frac{1}{36} ln ∣ x + 3 ∣ - \frac{1}{196} ln ∣ x + 7 ∣

= \frac{x}{x ^{2} + 10 x + 21} + 8 (\frac{1}{32} ln ∣ 2 x + 6 ∣ + \frac{3}{16 ( 2 x + 6 )} - \frac{1}{32} ln ∣ 2 x + 14 ∣ - \frac{7}{16 ( 2 x + 14 )}) - \frac{10}{441} ln ∣ x ∣ - \frac{1}{21 x} + \frac{1}{36} ln ∣ x + 3 ∣ - \frac{1}{196} ln ∣ x + 7 ∣

Add a constant to the solution

= \frac{x}{x ^{2} + 10 x + 21} + 8 (\frac{1}{32} ln ∣ 2 x + 6 ∣ + \frac{3}{16 ( 2 x + 6 )} - \frac{1}{32} ln ∣ 2 x + 14 ∣ - \frac{7}{16 ( 2 x + 14 )}) - \frac{10}{441} ln ∣ x ∣ - \frac{1}{21 x} + \frac{1}{36} ln ∣ x + 3 ∣ - \frac{1}{196} ln ∣ x + 7 ∣ + C

\int + arctan (7 x) d x

\int \frac{x}{16 + x ^{4}} d x

in t e g r a l x^{2} e^{x^{3}}

\int x (x + 8)^{4} d x

\int \frac{v}{2 v ^{2} + 1} d v

What is the integral of (x^2+1)/(x^4+10x^3+21x^2) ?
The integral of (x^2+1)/(x^4+10x^3+21x^2) is