What is the integral of (ln(p+1))/(p+1) ?

The integral of (ln(p+1))/(p+1) is (ln^2(1+p))/2+C

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integral of (ln(p+1))/(p+1)

\int \frac{ln ( p + 1 )}{p + 1} d p

\frac{ln ^{2} ( 1 + p )}{2} + C

Solution steps

\int \frac{ln ( p + 1 )}{p + 1} d p

\frac{ln ( p + 1 )}{p + 1} = ln (p + 1) \frac{1}{p + 1}

= \int ln (p + 1) \frac{1}{p + 1} d p

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = ln (1 + p)

= \frac{ln ^{2} ( 1 + p )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( 1 + p )}{2} + C

x \to 0 lim (\frac{1}{1 + e ^{\frac{1}{x}}})

\frac{d}{d x} (arcsin (\frac{x}{x + y}))

y^{^{'}} = \frac{( x ^{2} + y ^{2} )}{( 2 x y )}

f (x) = ∣ 2 x - 1 ∣

y^{^{'}} + 5 y = 6 e^{x}

What is the integral of (ln(p+1))/(p+1) ?
The integral of (ln(p+1))/(p+1) is (ln^2(1+p))/2+C