Primitive of x over a x + b squared

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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 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:

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

$\blacksquare$


Proof 2

\(\ds \int \frac {x \rd x} {\paren {a x + b}^2}\) \(=\) \(\ds \int \frac {a x \rd x} {a \paren {a x + b}^2}\) 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

$\blacksquare$


Also presented as

This result is also seen presented in the form:

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


Also see


Sources

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