What is the integral from 0 to 1 of z/((z^2+1)^3) ?

The integral from 0 to 1 of z/((z^2+1)^3) is 3/16

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integral from 0 to 1 of z/((z^2+1)^3)

\int_{0}^{1} \frac{z}{( z ^{2} + 1 ) ^{3}} d z

\frac{3}{16}

Decimal Notation

0.1875

Solution steps

\int_{0}^{1} \frac{z}{( z ^{2} + 1 ) ^{3}} d z

Apply u-substitution

: \int_{1}^{2} \frac{1}{2 u ^{3}} d u

= \int_{1}^{2} \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{1}^{2}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{1}^{2}

Compute the boundaries:

\frac{3}{8}

= \frac{1}{2} \cdot \frac{3}{8}

Simplify

= \frac{3}{16}

x \to 0 + lim (ln (x^{3}))

(1 + t^{2}) y^{^{'}} + 4 t y = t

x \to - 4 lim (\frac{1}{( x + 4 ) ^{4}})

\frac{d}{d x} (10 x + 4 y)

8 tan (2 x)

What is the integral from 0 to 1 of z/((z^2+1)^3) ?
The integral from 0 to 1 of z/((z^2+1)^3) is 3/16