What is the integral of (4+x^7)^37x^6 ?

The integral of (4+x^7)^37x^6 is 1/4 (4+x^7)^4+C

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integral of (4+x^7)^37x^6

\int (4 + x^{7})^{3} 7 x^{6} d x

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

Solution steps

\int (4 + x^{7})^{3} \cdot 7 x^{6} d x

Take the constant out:

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

= 7 \cdot \int (4 + x^{7})^{3} x^{6} d x

Apply u-substitution

: \int \frac{u ^{3}}{7} d u

= 7 \cdot \int \frac{u ^{3}}{7} d u

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = 4 + x^{7}

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

Simplify

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

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

Add a constant to the solution

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

5 x^{2} cos (3 x)

y^{^{''}} - \frac{7}{t} y^{^{'}} + \frac{32}{t ^{2}} y = 0

x^{2} \cdot y^{^{'}} - 4 x y = sin (3 x) \cdot x^{6}

x \to 0 lim (\frac{sin ( x ^{5} )}{x})

\int \frac{1}{1 + 9 x ^{2}} d x

What is the integral of (4+x^7)^37x^6 ?
The integral of (4+x^7)^37x^6 is 1/4 (4+x^7)^4+C