Definite integral x/(1+√(x+1)) from −1 to 3

This problem asks to evaluate the definite integral of an irrational function on the interval [−1, 3].

Problem

$\int_{- 1}^{3} \frac{x}{1 + \sqrt{x + 1}} d x$

Solution

Step 1

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

Why is the substitution t = √(x+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 = (2t/a) dt also contains no roots. As a result, the integral becomes an integral of a rational function.

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

Step 2

Find the new limits of integration:

How do we recalculate the limits of integration?

When substituting a variable, the limits of integration are recalculated. Substituting the old limits into t = √(x+1): for x = −1 we get t = √0 = 0; for x = 3 we get t = √4 = 2. So the old interval [−1, 3] becomes the new interval [0, 2].

x = - 1 \Rightarrow t = 0, x = 3 \Rightarrow t = 2

Step 3

Substitute the change of variable into the integral:

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

Step 4

Take out the factor 2 and reduce the integrand to the form:

How do we simplify the integrand?

The integrand has the form t(t²−1)/(t+1). The numerator t(t²−1) factors as t(t−1)(t+1). Then the fraction becomes t(t−1)(t+1)/(t+1), and the factor (t+1) cancels. As a result, we obtain the simple function t(t−1) = t² − t, easily integrated term by term.

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

Step 5

Factor the numerator and cancel:

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

Step 6

Integrate term by term: ∫ t² dt = t³/3, ∫ t dt = t²/2. Apply the Newton-Leibniz formula:

2 \int_{0}^{2} (t^{2} - t) d t = 2 [_{0}^{2} \frac{t^{3}}{3} - \frac{t^{2}}{2}

Step 7

Substitute the limits t = 2 and t = 0. After simplification, we obtain the answer.

Answer

$\frac{4}{3}$

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.

How do we find the new limits of integration?

Substitute the old limits in x into t = √(x+1). For the lower limit x = −1: t = √0 = 0. For the upper limit x = 3: t = √4 = 2. We get new limits [0, 2].

Why does the factor (t+1) cancel?

The numerator t(t²−1) = t(t−1)(t+1) contains the factor (t+1), which coincides with the denominator. After cancellation, t(t−1) = t² − t remains. Cancellation is valid because on [0, 2] the denominator t+1 ≠ 0 (it equals zero only at t = −1, outside the interval).

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 (√(x+1)), so the rational substitution t = √(x+1) is sufficient.