Primitive of Reciprocal of x by a x + b squared

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Theorem

$\ds \int \frac {\d x} {x \paren {a x + b}^2} = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C$


Proof 1

\(\ds \int \frac {\d x} {x \paren {a x + b}^2}\) \(=\) \(\ds \int \paren {\frac 1 {b^2 x} - \frac a {b^2 \paren {a x + b} } - \frac a {b \paren {a x + b}^2} } \rd x\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \frac 1 {b^2} \int \frac {\d x} x - \frac a {b^2} \int \frac {\d x} {a x + b} - \frac a b \int \frac {\d x} {\paren {a x + b}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {b^2} \ln \size x - \frac a {b^2} \int \frac {\d x} {a x + b} - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac 1 {b^2} \ln \size x - \frac a {b^2} \ln \size {a x + b} - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C\) Primitive of $\dfrac 1 {a x + b}$
\(\ds \) \(=\) \(\ds \frac 1 {b^2} \ln \size x - \frac a {b^2} \ln \size {a x + b} - \frac a b \frac {-1} {a \paren {a x + b} } + C\) Primitive of $\dfrac 1 {\paren {a x + b}^2}$
\(\ds \) \(=\) \(\ds \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C\) Difference of Logarithms and rearranging

$\blacksquare$


Proof 2

\(\ds \int \frac {\d x} {x \paren {a x + b}^2}\) \(=\) \(\ds \int \frac {b \rd x} {b x \paren {a x + b}^2}\) multiplying top and bottom by $b$
\(\ds \) \(=\) \(\ds \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b}^2}\) adding and subtracting $a x$
\(\ds \) \(=\) \(\ds \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b}^2} - \frac a b \int \frac {x \rd x} {x \paren {a x + b}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 b \int \frac {\d x} {x \paren {a x + b} } - \frac a b \int \frac {\d x} {\paren {a x + b}^2}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C\) Primitive of $\dfrac 1 {x \paren {a x + b} }$
\(\ds \) \(=\) \(\ds \frac 1 {b^2} \ln \size {\frac x {a x + b} } - \frac a b \paren {-\frac 1 {a \paren {a x + b} } } + C\) Primitive of $\dfrac 1 {\paren {a x + b}^2}$
\(\ds \) \(=\) \(\ds \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C\) simplifying

$\blacksquare$


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