Primitive of x over a x + b squared/Proof 2
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Theorem
- $\ds \int \frac {x \rd x} {\paren {a x + b}^2} = \frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} + C$
Proof
\(\ds \int \frac {x \rd x} {\paren {a x + b}^2}\) | \(=\) | \(\ds \int \frac {a x \rd x} {a \paren {a x + b}^2}\) | multiplying top and bottom by $a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\paren {a x + b - b} \rd x} {a \paren {a x + b}^2}\) | adding and subtracting $b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\paren {a x + b} \rd x} {\paren {a x + b}^2} - \frac b a \int \frac {\d x} {\paren {a x + b}^2}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\d x} {a x + b} - \frac b a \int \frac {\d x} {\paren {a x + b}^2}\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \paren {\frac 1 a \ln \size {a x + b} } - \frac b a \int \frac {\d x} {\paren {a x + b}^2} + C\) | Primitive of Reciprocal of $\dfrac 1 {a x + b}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \paren {\frac 1 a \ln \size {a x + b} } - \frac b a \paren {-\frac 1 {a \paren {a x + b} } } + C\) | Primitive of Reciprocal of $\dfrac 1 {\paren {a x + b}^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} + C\) | simplification |
$\blacksquare$