What is the integral of 2y(5x-y^2)^{15} ?

The integral of 2y(5x-y^2)^{15} is -1/16 (-y^2+5x)^{16}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2y(5x-y^2)^{15}

\int 2 y (5 x - y^{2})^{15} d y

- \frac{1}{16} (- y^{2} + 5 x)^{16} + C

Solution steps

\int 2 y (5 x - y^{2})^{15} d y

Take the constant out:

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

= 2 \cdot \int (- y^{2} + 5 x)^{15} y d y

Apply u-substitution

: \int - \frac{u ^{15}}{2} d u

= 2 \cdot \int - \frac{u ^{15}}{2} d u

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int u^{15} d u)

Apply the Power Rule

: \frac{u ^{16}}{16}

= 2 (- \frac{1}{2} \cdot \frac{u ^{16}}{16})

Substitute back

u = - y^{2} + 5 x

= 2 (- \frac{1}{2} \cdot \frac{( - y ^{2} + 5 x ) ^{16}}{16})

Simplify

2 (- \frac{1}{2} \cdot \frac{( - y ^{2} + 5 x ) ^{16}}{16}) : - \frac{1}{16} (- y^{2} + 5 x)^{16}

= - \frac{1}{16} (- y^{2} + 5 x)^{16}

Add a constant to the solution

= - \frac{1}{16} (- y^{2} + 5 x)^{16} + C

\int \frac{9}{5 x 49 - 25 x ^{2}} d x

\int \frac{3 sin ( x )}{1 + cos ^{2} ( x )} d x

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

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

\int \frac{t ^{2}}{4 t - t ^{2}} d t

What is the integral of 2y(5x-y^2)^{15} ?
The integral of 2y(5x-y^2)^{15} is -1/16 (-y^2+5x)^{16}+C