What is the integral of (1+4x^2)^{1/2} ?

The integral of (1+4x^2)^{1/2} is x(1+4x^2)^{1/2}-1/4 (-ln|2x+sqrt(4x^2+1)|+2xsqrt(4x^2+1))+C

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integral of (1+4x^2)^{1/2}

\int (1 + 4 x^{2})^{\frac{1}{2}} d x

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

Solution steps

\int (1 + 4 x^{2})^{\frac{1}{2}} d x

Apply Integration By Parts

: x (1 + 4 x^{2})^{\frac{1}{2}} - \int \frac{4 x ^{2}}{( 1 + 4 x ^{2} ) ^{\frac{1}{2}}} d x

= x (1 + 4 x^{2})^{\frac{1}{2}} - \int \frac{4 x ^{2}}{( 1 + 4 x ^{2} ) ^{\frac{1}{2}}} d x

\int \frac{4 x ^{2}}{( 1 + 4 x ^{2} ) ^{\frac{1}{2}}} d x = \frac{1}{4} (- ln 2 x + 4 x^{2} + 1 + 2 x 4 x^{2} + 1)

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

Add a constant to the solution

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

2 (ln (y)) y^{^{'}} - x^{3} y = 0

\int (- 3) d x

\int e^{s i n (θ)} d θ

f (x) = 4 + 3 x - 2 x^{2}, at x = 2

y = x (x^{3} + 3)^{5}

What is the integral of (1+4x^2)^{1/2} ?
The integral of (1+4x^2)^{1/2} is x(1+4x^2)^{1/2}-1/4 (-ln|2x+sqrt(4x^2+1)|+2xsqrt(4x^2+1))+C