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

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

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integral of (x^2+1)/(x^3+x^2)

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

- ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 1}{x ^{3} + x ^{2}} : - \frac{1}{x} + \frac{1}{x ^{2}} + \frac{2}{x + 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

= - ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ + C

y e^{x}

\int \frac{1}{9 + ( x - 1 ) ^{2}} d x

f (x) = 3 x^{2}, (- 2, 12)

\int 8 x^{6} y^{3} + 15 x^{5} y^{4} d y

\frac{d}{d x} (x^{4} e^{2 x})

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