What is the integral of (ax^2-b)^5x ?

The integral of (ax^2-b)^5x is ((ax^2-b)^6)/(12a)+C

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integral of (ax^2-b)^5x

\int (a x^{2} - b)^{5} x d x

\frac{( a x ^{2} - b ) ^{6}}{12 a} + C

Solution steps

\int (a x^{2} - b)^{5} x d x

Apply u-substitution

= \int \frac{u ^{5}}{2 a} d u

Take the constant out:

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

= \frac{1}{2 a} \cdot \int u^{5} d u

Apply the Power Rule

= \frac{1}{2 a} \cdot \frac{u ^{6}}{6}

Substitute back

u = a x^{2} - b

= \frac{1}{2 a} \cdot \frac{( a x ^{2} - b ) ^{6}}{6}

Simplify

\frac{1}{2 a} \cdot \frac{( a x ^{2} - b ) ^{6}}{6} : \frac{( a x ^{2} - b ) ^{6}}{12 a}

= \frac{( a x ^{2} - b ) ^{6}}{12 a}

Add a constant to the solution

= \frac{( a x ^{2} - b ) ^{6}}{12 a} + C

What is the integral of (ax^2-b)^5x ?
The integral of (ax^2-b)^5x is ((ax^2-b)^6)/(12a)+C