What is the integral of (2x-1)^6 ?

The integral of (2x-1)^6 is 1/14 (2x-1)^7+C

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integral of (2x-1)^6

\int (2 x - 1)^{6} d x

\frac{1}{14} (2 x - 1)^{7} + C

Solution steps

\int (2 x - 1)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{2} d u

= \int \frac{u ^{6}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{1}{2} \cdot \frac{u ^{7}}{7}

Substitute back

u = 2 x - 1

= \frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{7}}{7}

Simplify

\frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{7}}{7} : \frac{1}{14} (2 x - 1)^{7}

= \frac{1}{14} (2 x - 1)^{7}

Add a constant to the solution

= \frac{1}{14} (2 x - 1)^{7} + C

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

y = x^{2} - 9 x, (7, - 14)

\int_{- 1}^{1} 3 x d x

x \to 0 + lim (x^{15} ln (x))

f (x) = \frac{1 - 3 t}{2 + t}