Question

Calculate each of the integrals in Exercises 17–46. For some integrals you may need to use polynomial long division, partial fractions, factoring or expanding, or the method of completing the square.

$\int \frac{2x^2+4x}{x^3+x^2+x+1}dx$

Short answer

The value of integral is$\frac{3}{2}\ln \left(\left|x^2+1\right|\right)+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right)+C$.

Step-by-step solution

Step 1. Given Information.

The given integral is$\int \frac{2x^2+4x}{x^3+x^2+x+1}dx$.

Step 2. Calculation.

Rewrite the integral:

$\int \frac{2x^2+4x}{x^3+x^2+x+1}dx=2\int \frac{x\left(x+2\right)}{x^3+x^2+x+1}dx$

The partial fraction written as:

$\frac{x\left(x+2\right)}{x^3+x^2+x+1}=\frac{3x+1}{2x^2+2}-\frac{1}{2x+2}$

Now we solve the integral as:

$$\begin{gathered} 2\int \frac{x\left(x+2\right)}{x^3+x^2+x+1}dx \\ =2\int \left(\frac{3x+1}{2x^2+2}-\frac{1}{2x+2}\right)dx \\ =2\int \frac{3x+1}{2x^2+2}dx-2\int \frac{1}{2x+2}dx \\ =2\int \frac{3x+1}{2\left(x^2+1\right)}dx-2\int \frac{1}{2\left(x+1\right)}dx \\ =\int \frac{3x+1}{\left(x^2+1\right)}dx-\int \frac{1}{\left(x+1\right)}dx \end{gathered}$$

Step 3. Calculation.

$\int \frac{3x+1}{\left(x^2+1\right)}dx-\int \frac{1}{\left(x+1\right)}dx$

Now let $u=x+1,du=dx$

role="math" localid="1649665272994" $$\begin{gathered} =\int \frac{3x+1}{\left(x^2+1\right)}dx-\int \frac{1}{u}du \\ =\int \frac{3x+1}{\left(x^2+1\right)}dx-\ln \left(\left|u\right|\right) \\ =\int \frac{3x+1}{\left(x^2+1\right)}dx-\ln \left(\left|x+1\right|\right) \\ =\int \frac{3x}{\left(x^2+1\right)}dx+\int \frac{1}{\left(x^2+1\right)}dx-\ln \left(\left|x+1\right|\right) \\ =\int \frac{3x}{\left(x^2+1\right)}dx+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right) \end{gathered}$$

Let $u=x^2+1,du=2xdx$

role="math" localid="1649665691698" $$\begin{gathered} \int \frac{3x}{\left(x^2+1\right)}dx+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right) \\ =\frac{3}{2}\int \frac{1}{u}du+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right)+C \\ =\frac{3}{2}\ln \left(\left|u\right|\right)+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right)+C \\ =\frac{3}{2}\ln \left(\left|x^2+1\right|\right)+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right)+C \end{gathered}$$

Step 4. Conclusion.

The value of integral is$\frac{3}{2}\ln \left(\left|x^2+1\right|\right)+{\tan }^{-1}\left(x\right)-\ln \left(\left|x+1\right|\right)+C$.