Definite integral x/(2+√(2x+1)) from −1/2 to 0

This problem asks to evaluate the definite integral of an irrational function on the interval [−1/2, 0].

Problem

$\int_{- 1 / 2}^{0} \frac{x}{2 + \sqrt{2 x + 1}} d x$

Solution

Step 1

The integrand contains a square root. Apply the substitution t = √(2x+1). Then t² = 2x + 1, hence x = (t²−1)/2 and dx = t dt:

Why is the substitution t = √(2x+1) chosen?

If the integrand contains √(ax+b), the standard substitution t = √(ax+b) eliminates the irrationality. After squaring, we get t² = ax+b, hence x is expressed in terms of t as a rational function, and dx = t dt also contains no roots. As a result, the integral becomes an integral of a rational function, which is evaluated by standard methods.

t = \sqrt{2 x + 1}, t^{2} = 2 x + 1, x = \frac{t^{2} - 1}{2}, d x = t d t

Step 2

Find the new limits of integration:

x = - \frac{1}{2} \Rightarrow t = 0, x = 0 \Rightarrow t = 1

Step 3

Substitute the change of variable into the integral:

\int_{0}^{1} \frac{t^{2} - 1}{2} \cdot \frac{1}{2 + t} \cdot t d t

Step 4

Take out the factor 1/2 and reduce the integrand to a rational fraction:

Why do we divide the polynomial by the polynomial?

The integrand has the form t(t²−1)/(t+2) — a rational fraction. The degree of the numerator (3) is greater than the degree of the denominator (1), so the fraction is improper. To integrate it, we extract the polynomial part by dividing the numerator by the denominator: t(t²−1) = (t+2)(t²−2t+3) − 6. As a result, the integral splits into a sum of simple terms and one partial fraction 6/(t+2).

\frac{1}{2} \int_{0}^{1} \frac{t (t^{2} - 1)}{t + 2} d t

Step 5

Divide the numerator t(t²−1) = t³ − t by (t+2):

\frac{t (t^{2} - 1)}{t + 2} = t^{2} - 2 t + 3 - \frac{6}{t + 2}

Step 6

Substitute the decomposition into the integral:

\frac{1}{2} \int_{0}^{1} (t^{2} - 2 t + 3 - \frac{6}{t + 2}) d t

Step 7

Integrate term by term: ∫ t² dt = t³/3, ∫ 2t dt = t², ∫ 3 dt = 3t, ∫ 6/(t+2) dt = 6 ln|t+2|. Apply the Newton-Leibniz formula:

\frac{1}{2} [_{0}^{1} \frac{t^{3}}{3} - t^{2} + 3 t - 6 ln | t + 2 |

Step 8

Substitute the limits t = 1 and t = 0. After simplification, we obtain the answer.

Answer

$\frac{7}{6} - 3 ln \frac{3}{2}$

Frequently asked questions

How do I evaluate an integral with a square root in the denominator?

If the integrand contains √(ax+b), apply the substitution t = √(ax+b). Then t² = ax+b, x = (t²−b)/a, dx = (2t/a) dt. This eliminates the root and turns the integral into a rational function, which is then integrated by extracting the polynomial part and decomposing into partial fractions.

How do we find the new limits of integration?

Substitute the old limits in x into t = √(2x+1). For the lower limit x = −1/2: t = √0 = 0. For the upper limit x = 0: t = √1 = 1. We get new limits [0, 1].

Why does a logarithm appear in the answer?

After dividing the polynomial t³ − t by (t+2), the remainder is −6, and one of the integrals has the form ∫ 6/(t+2) dt = 6 ln|t+2|. Substituting the limits t = 1 and t = 0 gives 6(ln 3 − ln 2) = 6 ln(3/2), which after multiplying by 1/2 gives 3 ln(3/2).

What about irrationalities of the form √(ax²+bx+c)?

If the root has the form √(ax²+bx+c), trigonometric or hyperbolic substitutions are used. For example, for √(a²−x²) — substitution x = a sin t; for √(a²+x²) — x = a tg t; for √(x²−a²) — x = a/cos t. In our case the root is linear (√(2x+1)), so the rational substitution t = √(2x+1) is sufficient.