What is the integral of (4x-3)^6 ?

The integral of (4x-3)^6 is 1/28 (4x-3)^7+C

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integral of (4x-3)^6

\int (4 x - 3)^{6} d x

\frac{1}{28} (4 x - 3)^{7} + C

Solution steps

\int (4 x - 3)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{4} d u

= \int \frac{u ^{6}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{6} d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{u ^{7}}{7}

Substitute back

u = 4 x - 3

= \frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{7}}{7}

Simplify

\frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{7}}{7} : \frac{1}{28} (4 x - 3)^{7}

= \frac{1}{28} (4 x - 3)^{7}

Add a constant to the solution

= \frac{1}{28} (4 x - 3)^{7} + C

x \to 2 + lim (\frac{x + 3}{x - 2})

f (x) = \frac{6}{x}, at x = 1

x \to 2 lim (x^{2} - \frac{1}{x})

\frac{1}{s ^{2} ( s + 1 ) ^{2}}

\int \frac{x ^{5}}{( 3 - x ^{6} ) ^{2}} d x