Primitive of x squared over Cube of Root of a x squared plus b x plus c

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\ds \int \frac {x^2 \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {\paren {2 b^2 - 4 a c} x + 2 b c} {a \paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$


Proof

\(\ds \) \(\) \(\ds \int \frac {x^2 \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}\)
\(\ds \) \(=\) \(\ds \int \frac {a x^2 \rd x} {a \paren {\sqrt {a x^2 + b x + c} }^3}\) multiplying top and bottom by $a$
\(\ds \) \(=\) \(\ds \int \frac {\paren {a x^2 + b x + c - b x - c} \rd x} {a \paren {\sqrt {a x^2 + b x + c} }^3}\)
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\paren {a x^2 + b x + c} \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b a \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac c a \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac b a \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac c a \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b a \paren {\frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } }\) Primitive of $\dfrac x {\paren {\sqrt {a x^2 + b x + c} }^3}$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac c a \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }\) Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
\(\ds \) \(=\) \(\ds \frac {\paren {2 b^2 - 4 a c} x + 2 b c} {a \paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + \frac 1 a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) simplifying

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_x_squared_over_Cube_of_Root_of_a_x_squared_plus_b_x_plus_c&oldid=598226"