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Popular >

integral of 1/(1+y)

Solution

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

Solution

ln ∣ 1 + y ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 1 + y

= ln ∣ 1 + y ∣

Add a constant to the solution

= ln ∣ 1 + y ∣ + C

Popular Examples

integral of xsqrt(9+x^2)

\int x 9 + x^{2} d x

derivative of y= 1/4 x^2-1/2 ln(x)derivative of

y = \frac{1}{4} x^{2} - \frac{1}{2} ln (x)

sum from n=1 to infinity of 1/(e^{n^2)}

n = 1 \sum \infty \frac{1}{e ^{n^{2}}}

(\partial)/(\partial y)(2xe^y)

\frac{\partial}{\partial y} (2 x e^{y})

tangent of 15-x^2 tangent of

15 - x^{2}

Frequently Asked Questions (FAQ)

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