What is the integral of (y^3e^{y^2})/((y^2+1)^2) ?

The integral of (y^3e^{y^2})/((y^2+1)^2) is (e^{y^2})/(2(y^2+1))+C

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integral of (y^3e^{y^2})/((y^2+1)^2)

\int \frac{y ^{3} e ^{y^{2}}}{( y ^{2} + 1 ) ^{2}} d y

\frac{e ^{y^{2}}}{2 ( y ^{2} + 1 )} + C

Solution steps

\int \frac{y ^{3} e ^{y^{2}}}{( y ^{2} + 1 ) ^{2}} d y

Apply u-substitution

: \int \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3 ( u ^{\frac{2}{3}} + 1 ) ^{2}} d u

= \int \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3 ( u ^{\frac{2}{3}} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{( u ^{\frac{2}{3}} + 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{3 e ^{v} v}{2 ( v + 1 ) ^{2}} d v

= \frac{1}{3} \cdot \int \frac{3 e ^{v} v}{2 ( v + 1 ) ^{2}} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int \frac{e ^{v} v}{( v + 1 ) ^{2}} d v

Apply Integration By Parts

: - \frac{e ^{v} v}{v + 1} - \int - e^{v} d v

= \frac{1}{3} \cdot \frac{3}{2} (- \frac{e ^{v} v}{v + 1} - \int - e^{v} d v)

\int - e^{v} d v = - e^{v}

= \frac{1}{3} \cdot \frac{3}{2} (- \frac{e ^{v} v}{v + 1} - (- e^{v}))

Substitute back

= \frac{1}{3} \cdot \frac{3}{2} - \frac{e ^{(y^{3})^{\frac{2}{3}}} ( y ^{3} ) ^{\frac{2}{3}}}{( y ^{3} ) ^{\frac{2}{3}} + 1} - (- e^{(y^{3})^{\frac{2}{3}}})

Simplify

\frac{1}{3} \cdot \frac{3}{2} - \frac{e ^{(y^{3})^{\frac{2}{3}}} ( y ^{3} ) ^{\frac{2}{3}}}{( y ^{3} ) ^{\frac{2}{3}} + 1} - (- e^{(y^{3})^{\frac{2}{3}}}) : \frac{e ^{y^{2}}}{2 ( y ^{2} + 1 )}

= \frac{e ^{y^{2}}}{2 ( y ^{2} + 1 )}

Add a constant to the solution

= \frac{e ^{y^{2}}}{2 ( y ^{2} + 1 )} + C

f (x) = x^{2} - 2 x, (0, 2)

y^{^{''}} - 8 y^{^{'}} + 15 y = 4 e^{t}

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

\frac{d}{d x} (lo g_{3} (4 x^{2} + 2 x))

What is the integral of (y^3e^{y^2})/((y^2+1)^2) ?
The integral of (y^3e^{y^2})/((y^2+1)^2) is (e^{y^2})/(2(y^2+1))+C