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Popular >

integral of y/((1+y^2))

Solution

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

Solution

\frac{1}{2} ln 1 + y^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} 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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + y^{2}

= \frac{1}{2} ln 1 + y^{2}

Add a constant to the solution

= \frac{1}{2} ln 1 + y^{2} + C

Popular Examples

integral of (4x^2-3x+4)/(x^3+4x)

\int \frac{4 x ^{2} - 3 x + 4}{x ^{3} + 4 x} d x

y^{''}+2y^'+y=2e^{-x}

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- x}

derivative of 36x^2-72x

\frac{d}{d x} (36 x^{2} - 72 x)

limit as x approaches 0 of cos(pi/x)

x \to 0 lim (cos (\frac{π}{x}))

integral of (ln(x))/(x^{14)}

\int \frac{ln ( x )}{x ^{14}} d x

Frequently Asked Questions (FAQ)

What is the integral of y/((1+y^2)) ?
The integral of y/((1+y^2)) is 1/2 ln|1+y^2|+C