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

The integral of (x^2+1)/(x^3+3x^2-4) is 2/9 ln|x-1|+7/9 ln|x+2|+5/(3(x+2))+C

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

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

\frac{2}{9} ln ∣ x - 1 ∣ + \frac{7}{9} ln ∣ x + 2 ∣ + \frac{5}{3 ( x + 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 1}{x ^{3} + 3 x ^{2} - 4} : \frac{2}{9 ( x - 1 )} + \frac{7}{9 ( x + 2 )} - \frac{5}{3 ( x + 2 ) ^{2}}

= \int \frac{2}{9 ( x - 1 )} + \frac{7}{9 ( x + 2 )} - \frac{5}{3 ( 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{2}{9 ( x - 1 )} d x + \int \frac{7}{9 ( x + 2 )} d x - \int \frac{5}{3 ( x + 2 ) ^{2}} d x

\int \frac{2}{9 ( x - 1 )} d x = \frac{2}{9} ln ∣ x - 1 ∣

\int \frac{7}{9 ( x + 2 )} d x = \frac{7}{9} ln ∣ x + 2 ∣

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

= \frac{2}{9} ln ∣ x - 1 ∣ + \frac{7}{9} ln ∣ x + 2 ∣ - (- \frac{5}{3 ( x + 2 )})

Simplify

= \frac{2}{9} ln ∣ x - 1 ∣ + \frac{7}{9} ln ∣ x + 2 ∣ + \frac{5}{3 ( x + 2 )}

Add a constant to the solution

= \frac{2}{9} ln ∣ x - 1 ∣ + \frac{7}{9} ln ∣ x + 2 ∣ + \frac{5}{3 ( x + 2 )} + C

x \to - 2 lim (x + 1)

y^{^{'}} = e^{9 x} - 8 x

\frac{60}{0.01 x ^{2} + 1}

\frac{\partial}{\partial x} (y^{\frac{1}{2}} + x^{\frac{1}{2}})

f (x) = 8 x^{2} + 5 x, at 53

What is the integral of (x^2+1)/(x^3+3x^2-4) ?
The integral of (x^2+1)/(x^3+3x^2-4) is 2/9 ln|x-1|+7/9 ln|x+2|+5/(3(x+2))+C