What is the integral of (1+2x^2)^2 ?

The integral of (1+2x^2)^2 is x+(4x^3)/3+(4x^5)/5+C

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

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

x + \frac{4 x ^{3}}{3} + \frac{4 x ^{5}}{5} + C

Solution steps

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

Expand

(1 + 2 x^{2})^{2} : 1 + 4 x^{2} + 4 x^{4}

= \int 1 + 4 x^{2} + 4 x^{4} d x

Apply the Sum Rule:

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

= \int 1 d x + \int 4 x^{2} d x + \int 4 x^{4} d x

\int 1 d x = x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 4 x^{4} d x = \frac{4 x ^{5}}{5}

= x + \frac{4 x ^{3}}{3} + \frac{4 x ^{5}}{5}

Add a constant to the solution

= x + \frac{4 x ^{3}}{3} + \frac{4 x ^{5}}{5} + C

\frac{d y}{d x} = \frac{2 x}{1 + 2 y}, y (2) = 1

\frac{2 x}{y} d x - \frac{x ^{2}}{y ^{2}} d y = 0

f (x) = \frac{2 x}{7} + 1

\frac{d y}{d x} = (x - 2) e^{- 2 y}

What is the integral of (1+2x^2)^2 ?
The integral of (1+2x^2)^2 is x+(4x^3)/3+(4x^5)/5+C