What is the integral of 8x(4x^2+3)^6 ?

The integral of 8x(4x^2+3)^6 is 1/7 (4x^2+3)^7+C

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integral of 8x(4x^2+3)^6

\int 8 x (4 x^{2} + 3)^{6} d x

\frac{1}{7} (4 x^{2} + 3)^{7} + C

Solution steps

\int 8 x (4 x^{2} + 3)^{6} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} + 3)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{8} d u

= 8 \cdot \int \frac{u ^{6}}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int u^{6} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{8} \cdot \frac{u ^{7}}{7}

Substitute back

u = 4 x^{2} + 3

= 8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{7}}{7}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{7}}{7} : \frac{1}{7} (4 x^{2} + 3)^{7}

= \frac{1}{7} (4 x^{2} + 3)^{7}

Add a constant to the solution

= \frac{1}{7} (4 x^{2} + 3)^{7} + C

\int_{0}^{1} \frac{1}{4 - x ^{2}} d x

h (t) = (\frac{t ^{3}}{t ^{8} + 3})^{2}

x = y^{2} - 6 x, x = y - y^{2}

\frac{\partial}{\partial x} (7 e^{x - y})

(3 x^{2} + 4 x y) d x + (2 x^{2} + 2 y) d y = 0

What is the integral of 8x(4x^2+3)^6 ?
The integral of 8x(4x^2+3)^6 is 1/7 (4x^2+3)^7+C