What is the integral of ln(a+bx) ?

The integral of ln(a+bx) is a 1/b ln(a+bx)+xln(a+bx)-a/b-x+C

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integral of ln(a+bx)

\int ln (a + b x) d x

a \frac{1}{b} ln (a + b x) + x ln (a + b x) - \frac{a}{b} - x + C

Solution steps

\int ln (a + b x) d x

Apply u-substitution

: \int \frac{ln ( u )}{b} d u

= \int \frac{ln ( u )}{b} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{b} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{b} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{b} (u ln (u) - u)

Substitute back

u = a + b x

= \frac{1}{b} ((a + b x) ln (a + b x) - (a + b x))

Simplify

\frac{1}{b} ((a + b x) ln (a + b x) - (a + b x)) : a \frac{1}{b} ln (a + b x) + x ln (a + b x) - \frac{a}{b} - x

= a \frac{1}{b} ln (a + b x) + x ln (a + b x) - \frac{a}{b} - x

Add a constant to the solution

= a \frac{1}{b} ln (a + b x) + x ln (a + b x) - \frac{a}{b} - x + C

\frac{1}{( s - 1 ) ^{2} - 4}

(x^{2} + 2) y^{^{'}} + 2 x y = x^{3}

\int_{0}^{5} 4 x^{2} d x

\int 2 z^{2} d z

\int \frac{1}{3 - 6 x - x ^{2}} d x

What is the integral of ln(a+bx) ?
The integral of ln(a+bx) is a 1/b ln(a+bx)+xln(a+bx)-a/b-x+C