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

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

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

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

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

Solution steps

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

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 (x) + 1

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

Add a constant to the solution

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

\int ln (x^{2}) d x

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

i = 1 \sum \infty 8 (\frac{5}{6})^{i - 1}

(cos (x))^{2} π

y^{^{'}} = \frac{x - 3}{y + 5}

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