What is the integral of ((1+6ln(x))^{11})/x ?

The integral of ((1+6ln(x))^{11})/x is 1/72 (1+6ln(x))^{12}+C

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integral of ((1+6ln(x))^{11})/x

\int \frac{( 1 + 6 ln ( x ) ) ^{11}}{x} d x

\frac{1}{72} (1 + 6 ln (x))^{12} + C

Solution steps

\int \frac{( 1 + 6 ln ( x ) ) ^{11}}{x} d x

Apply u-substitution

: \int \frac{u ^{11}}{6} d u

= \int \frac{u ^{11}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{11} d u

Apply the Power Rule

: \frac{u ^{12}}{12}

= \frac{1}{6} \cdot \frac{u ^{12}}{12}

Substitute back

u = 1 + 6 ln (x)

= \frac{1}{6} \cdot \frac{( 1 + 6 ln ( x ) ) ^{12}}{12}

Simplify

\frac{1}{6} \cdot \frac{( 1 + 6 ln ( x ) ) ^{12}}{12} : \frac{1}{72} (1 + 6 ln (x))^{12}

= \frac{1}{72} (1 + 6 ln (x))^{12}

Add a constant to the solution

= \frac{1}{72} (1 + 6 ln (x))^{12} + C

\int x^{2} + 2 x y + y^{2} d x

\frac{d y}{d x} = \frac{1}{1 - x ^{2}} - \frac{1}{x}

x^{2} + 4 x + 6

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

2 t^{2} - t + 1

What is the integral of ((1+6ln(x))^{11})/x ?
The integral of ((1+6ln(x))^{11})/x is 1/72 (1+6ln(x))^{12}+C