What is the integral of 2/((2-y)^2)\sqrt[3]{(2-y)/(2+y)} ?

The integral of 2/((2-y)^2)\sqrt[3]{(2-y)/(2+y)} is (3(2+y)^{2/3})/(4(2-y)^{2/3)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/((2-y)^2)\sqrt[3]{(2-y)/(2+y)}

\frac{3 ( 2 + y ) ^{\frac{2}{3}}}{4 ( 2 - y ) ^{\frac{2}{3}}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

: \int - \frac{3}{4 u ^{3}} d u

= 2 \cdot \int - \frac{3}{4 u ^{3}} d u

Take the constant out:

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

= 2 (- \frac{3}{4} \cdot \int \frac{1}{u ^{3}} d u)

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 2 (- \frac{3}{4} (- \frac{1}{2 u ^{2}}))

Substitute back

Simplify

= \frac{3 ( 2 + y ) ^{\frac{2}{3}}}{4 ( 2 - y ) ^{\frac{2}{3}}}

Add a constant to the solution

= \frac{3 ( 2 + y ) ^{\frac{2}{3}}}{4 ( 2 - y ) ^{\frac{2}{3}}} + C

\int \frac{1}{x 25 x ^{2} - 9} d x

\int \frac{ln ( 1 + x ) - ln ( x )}{x ( 1 + x )} d x

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

\int 9 t^{4} + 16 t^{2} d t

\int t^{2} 1 + \frac{1}{t ^{2}} d t

What is the integral of 2/((2-y)^2)\sqrt[3]{(2-y)/(2+y)} ?
The integral of 2/((2-y)^2)\sqrt[3]{(2-y)/(2+y)} is (3(2+y)^{2/3})/(4(2-y)^{2/3)}+C