## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}+x+1}{(x+1) x^{3}} d x
$$
## Solution

$$
\int \frac{2 x^{3}+x+1}{(x+1) x^{3}} d x=
$$

We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{2 x^{3}+x+1}{(x+1) x^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x}+\frac{B_{2}}{x^{2}}+\frac{B_{3}}{x^{3}}= \\
& =\frac{A x^{3}+B_{1}(x+1) x^{2}+B_{2}(x+1) x+B_{3}(x+1)}{(x+1) x^{3}}= \\
& =\frac{A x^{3}+B_{1}\left(x^{3}+x^{2}\right)+B_{2}\left(x^{2}+x\right)+B_{3}(x+1)}{(x+1) x^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(B_{1}+B_{2}\right) x^{2}+\left(B_{2}+B_{3}\right) x+B_{3}}{(x+1) x^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=2 \\
B_{1}+B_{2}=0 \\
B_{2}+B_{3}=1 \\
B_{3}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=2 \\
B_{1}+B_{2}=0 \\
B_{2}=0 \\
B_{3}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=2 \\
B_{1}=0 \\
B_{2}=0 \\
B_{3}=1
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
A=2 \\
B_{1}=0 \\
B_{2}=0 \\
B_{3}=1
\end{array}\right. \\
& \frac{2 x^{3}+x+1}{(x+1) x^{3}}=\frac{2}{x+1}+\frac{1}{x^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{2 x^{3}+x+1}{(x+1) x^{3}} d x=\int\left(\frac{2}{x+1}+\frac{1}{x^{3}}\right) d x=2 \cdot \ln |x+1|-\frac{1}{2 x^{2}}+C
$$

Source — "http://pluspi.org/wiki/index.php/%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%E2%80%8B%25D0%259A%25D1%2583%25D0%25B7%25D0%25BD%25D0%25B5%25D1%2586%25D0%25BE%25D0%25B2_%25D0%2598%25D0%25BD%25D1%2582%25D0%25B5%25D0%25B3%25D1%2580%25D0%25B0%25D0%25BB%25D1%258B%25D0%2597%25D0%25B0%25D0%25B4%25D0%25B0%25D1%2587%25D0%25B0_6-16"

Categories: Kuznetsov's Problem Book Integrals Problem 6 | Integrals

- Last modified: 08:45, April 9, 2009.
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## Problem Kuznetsov Integrals 6-17

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