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Popular >

x^'= 1/(1+t^2)

Solution

x^{^{'}} = \frac{1}{1 + t ^{2}}

Solution

x = arctan (t) + c_{1}

Solution steps

x^{'} (t) = \frac{1}{1 + t ^{2}}

If

f^{^{'}} (x) = g (x)

then

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

x = \int \frac{1}{1 + t ^{2}} d t

\int \frac{1}{1 + t ^{2}} d t = arctan (t) + c_{1}

x = arctan (t) + c_{1}

Popular Examples

tangent of f(x)=4-2x^2,\at x=5 tangent of

f (x) = 4 - 2 x^{2}, at x = 5

derivative of ((-2x-1)/(-4x+2))^5 derivative of

(\frac{- 2 x - 1}{- 4 x + 2})^{5}

(\partial)/(\partial y)(x^2+xy^4)

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

integral from 1 to x of 1/(sqrt(t))

\int_{1}^{x} \frac{1}{t} d t

integral of 1/((1+sqrt(x))^2)

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

Frequently Asked Questions (FAQ)

What is the solution for x^'= 1/(1+t^2) ?
The solution for x^'= 1/(1+t^2) is x=arctan(t)+c_{1}