Primitive of Reciprocal of x by x cubed plus a cubed squared
Jump to navigation
Jump to search
Theorem
- $\ds \int \frac {\d x} {x \paren {x^3 + a^3}^2} = \frac 1 {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^6} \ln \size {\frac {x^3} {x^3 + a^3} }$
Proof
\(\ds \int \frac {\d x} {x \paren {x^3 + a^3}^2}\) | \(=\) | \(\ds \int \frac {a^3 \rd x} {a^3 x \paren {x^3 + a^3}^2}\) | multiplying top and bottom by $a^3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\paren {x^3 + a^3 - x^3} \rd x} {a^3 x \paren {x^3 + a^3}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^3} \int \frac {\paren {x^3 + a^3} \rd x} {x \paren {x^3 + a^3}^2} - \frac 1 {a^3} \int \frac {x^3 \rd x} {x \paren {x^3 + a^3}^2}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^3} \int \frac {\d x} {x \paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x^2 \rd x} {\paren {x^3 + a^3}^2}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^3} \paren {\frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } } - \frac 1 {a^3} \int \frac {x^2 \rd x} {\paren {x^3 + a^3}^2}\) | Primitive of $\dfrac 1 {x \paren {x^3 + a^3} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^3} \paren {\frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } } - \frac 1 {a^3} \paren {\frac {-1} {3 \paren {x^3 + a^3} } }\) | Primitive of $\dfrac {x^2} {\paren {x^3 + a^3}^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^6} \ln \size {\frac {x^3} {x^3 + a^3} }\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^3 + a^3$: $14.307$