Primitive of x squared over a x + b squared

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2} = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C$


Proof 1

Put $u = a x + b$.

Then:

\(\ds x\) \(=\) \(\ds \frac {u - b} a\)
\(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 a\)


Then:

\(\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2}\) \(=\) \(\ds \int \frac 1 a \paren {\frac {u - b} a}^2 \frac 1 {u^2} \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac 1 {a^3} \paren {1 - \frac {2 b} u + \frac {b^2} {u^2} } \rd u\) Square of Difference, and simplification
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \d u - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac u {a^3} - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \frac {-1} u + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C\) substituting for $u$ and rearranging

$\blacksquare$


Proof 2

\(\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2}\) \(=\) \(\ds \int \frac {a x^2 \rd x} {a \paren {a x + b}^2}\) multiplying top and bottom by $a$
\(\ds \) \(=\) \(\ds \int \frac {x \paren {a x + b - b} \rd x} {a \paren {a x + b}^2}\) adding and subtracting $b x$
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {x \paren {a x + b} \rd x} {\paren {a x + b}^2} - \frac b a \int \frac {x \rd x} {\paren {a x + b}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {x \rd x} {a x + b} - \frac b a \int \frac {x \rd x} {\paren {a x + b}^2}\) simplification
\(\ds \) \(=\) \(\ds \frac 1 a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } - \frac b a \int \frac {\d x} {\paren {a x + b}^2} + C\) Primitive of Reciprocal of $\dfrac x {\paren {a x + b} }$
\(\ds \) \(=\) \(\ds \frac 1 a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } - \frac b a \paren {\frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} } + C\) Primitive of Reciprocal of $\dfrac x {\paren {a x + b}^2}$
\(\ds \) \(=\) \(\ds \frac x {a^2} - \frac b {a^3} \ln \size {a x + b} - \frac {b^2} {a^3 \paren {a x + b} } - \frac b {a^3} \ln \size {a x + b} + C\) multiplying out
\(\ds \) \(=\) \(\ds \frac x {a^2} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C\) gathering terms
\(\ds \) \(=\) \(\ds \frac {a x} {a^3} + \frac b {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C\) where $\dfrac b {a^3}$ is subsumed into $C$
\(\ds \) \(=\) \(\ds \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C\) simplification

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_x_squared_over_a_x_%2B_b_squared&oldid=740350"