What is the integral of log_{4}(3x+1) ?

The integral of log_{4}(3x+1) is xlog_{4}(3x+1)-(3x+1-ln|3x+1|)/(6ln(2))+C

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integral of log_{4}(3x+1)

\int lo g_{4} (3 x + 1) d x

x lo g_{4} (3 x + 1) - \frac{3 x + 1 - ln ∣ 3 x + 1 ∣}{6 ln ( 2 )} + C

Solution steps

\int lo g_{4} (3 x + 1) d x

Apply Integration By Parts

: x lo g_{4} (3 x + 1) - \int \frac{3 x}{2 ln ( 2 ) ( 3 x + 1 )} d x

= x lo g_{4} (3 x + 1) - \int \frac{3 x}{2 ln ( 2 ) ( 3 x + 1 )} d x

\int \frac{3 x}{2 ln ( 2 ) ( 3 x + 1 )} d x = \frac{3 x + 1 - ln ∣ 3 x + 1 ∣}{6 ln ( 2 )}

= x lo g_{4} (3 x + 1) - \frac{3 x + 1 - ln ∣ 3 x + 1 ∣}{6 ln ( 2 )}

Add a constant to the solution

= x lo g_{4} (3 x + 1) - \frac{3 x + 1 - ln ∣ 3 x + 1 ∣}{6 ln ( 2 )} + C

3 y^{^{''}} + 2 y^{^{'}} - y = 0

\int x (2 x + 1) e^{4 x^{3} + 3 x^{2} - 4} d x

y^{^{'}} + y^{2} = 1

\frac{( 4 x ^{5} + 5 )}{x ^{3}}

x \frac{d y}{d x} + 6 y = 1 - \frac{4}{x}

What is the integral of log_{4}(3x+1) ?
The integral of log_{4}(3x+1) is xlog_{4}(3x+1)-(3x+1-ln|3x+1|)/(6ln(2))+C