What is the integral of 2x^3(t^4-3)^5 ?

The integral of 2x^3(t^4-3)^5 is (x^4(t^4-3)^5)/2+C

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integral of 2x^3(t^4-3)^5

\int 2 x^{3} (t^{4} - 3)^{5} d x

\frac{x ^{4} ( t ^{4} - 3 ) ^{5}}{2} + C

Solution steps

\int 2 x^{3} (t^{4} - 3)^{5} d x

Take the constant out:

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

= (- 3 + t^{4})^{5} \cdot 2 \cdot \int x^{3} d x

Apply the Power Rule

: \frac{x ^{4}}{4}

= (- 3 + t^{4})^{5} \cdot 2 \cdot \frac{x ^{4}}{4}

Simplify

(- 3 + t^{4})^{5} \cdot 2 \cdot \frac{x ^{4}}{4} : \frac{x ^{4} ( t ^{4} - 3 ) ^{5}}{2}

= \frac{x ^{4} ( t ^{4} - 3 ) ^{5}}{2}

Add a constant to the solution

= \frac{x ^{4} ( t ^{4} - 3 ) ^{5}}{2} + C

\frac{3}{( s + 1 ) ( s + 2 )}

\int \frac{14 x - 11}{( 2 x - 3 ) ( x + 1 )} d x

\int ln (y + 1) d y

\frac{s + 3}{4 s ^{2} + 5}

x \to 0 + lim (\frac{1}{x ^{13}})

What is the integral of 2x^3(t^4-3)^5 ?
The integral of 2x^3(t^4-3)^5 is (x^4(t^4-3)^5)/2+C