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Popular >

integral of (y+1)

Solution

\int (y + 1) d y

Solution

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

Solution steps

\int y + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y d y + \int 1 d y

\int y d y = \frac{y ^{2}}{2}

\int 1 d y = y

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

Add a constant to the solution

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

Popular Examples

(dy)/(dx)-(xy)/(1+x)=x

\frac{d y}{d x} - \frac{x y}{1 + x} = x

(\partial)/(\partial x)(ln(4x^2+9y^2+8))

\frac{\partial}{\partial x} (ln (4 x^{2} + 9 y^{2} + 8))

y^{^{''}} + 2 y = 1

y^{''}+2y=sin(xsqrt(2))

y^{^{''}} + 2 y = sin (x 2)

tangent of f(x)=4x^2+12x+7,\at x=-2 tangent of

f (x) = 4 x^{2} + 12 x + 7, at x = - 2

Frequently Asked Questions (FAQ)

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