Question

∫((1+∜x)/((1+√x)) solve this please.

Answer 1

Answer on Question #50472 – Math – Integral Calculus

Solve $J((1 + \sqrt[4]{x}) / ((1 + vx))dx$.

Solution

We’ll use a substitution $t = \sqrt[4]{x}$ to avoid radicals.

$$\begin{aligned} \int \frac{1 + \sqrt[4]{x}}{1 + \sqrt{x}} dx &= \int \frac{1 + \sqrt[4]{x}}{1 + \left(\sqrt[4]{x}\right)^2} dx = \left| \begin{array}{c} t = \sqrt[4]{x} \\ x = t^4 \\ dx = 4t^3 dt \end{array} \right| = \int \frac{1 + t}{1 + t^2} \cdot 4t^3 dt = 4\int \frac{t^4 + t^3}{t^2 + 1} dt \\ &= 4\int \frac{t^2(t^2 + 1) + t(t^2 + 1) - (t^2 + 1) - t + 1}{t^2 + 1} dt \\ &= 4\int \left(t^2 + t - 1 - \frac{t}{t^2 + 1} + \frac{1}{t^2 + 1}\right) dt \\ &= 4\int t^2 dt + 4\int t dt - 4\int dt - 4\int \frac{t}{t^2 + 1} dt + 4\int \frac{1}{t^2 + 1} dt \\ &= 4 \cdot \frac{t^3}{3} + 4 \cdot \frac{t^2}{2} - 4 \cdot t - 4\int \frac{t}{t^2 + 1} dt + 4\tan^{-1}t + C \\ &= \frac{4}{3}t^3 + 2t^2 - 4t + 4\tan^{-1}t + C - 2\int \frac{2t dt}{t^2 + 1} \\ &= \frac{4}{3}t^3 + 2t^2 - 4t + 4\tan^{-1}t + C - 2\int \frac{d(t^2 + 1)}{t^2 + 1} = |\text{substitution } u = t^2 + 1| \\ &= \frac{4}{3}t^3 + 2t^2 - 4t + 4\tan^{-1}t + C - 2\int \frac{du}{u} \\ &= \frac{4}{3}t^3 + 2t^2 - 4t + 4\tan^{-1}t - 2\ln|u| + C \\ &= \frac{4}{3}t^3 + 2t^2 - 4t + 4\tan^{-1}t - 2\ln|t^2 + 1| + C \\ &= \frac{4}{3}\sqrt[4]{x^3} + 2\sqrt{x} - 4\sqrt[4]{x} + 4\tan^{-1}\sqrt[4]{x} - 2\ln(\sqrt{x} + 1) + C \end{aligned}$$

where $C$ is an arbitrary real constant, the sign $|$ of absolute value is omitted, because expression $\sqrt{x} + 1$ is positive.

We used the next table integrals: $\int y^n dy = \frac{y^{n+1}}{n+1} + C$, $n \neq -1$, $\int \frac{dy}{y} = \ln|y| + C$,

$\int \frac{dy}{1 + y^2} = \tan^{-1}(y) + C$, where $C$ is an arbitrary real constant, $\tan^{-1}(y)$ means the inverse tangent function (it is sometimes called $\arctan(y)$).

Answer:

$$\int \frac{1 + \sqrt[4]{x}}{1 + \sqrt{x}} dx = \frac{4}{3}\sqrt[4]{x^3} + 2\sqrt{x} - 4\sqrt[4]{x} + 4\tan^{-1}\sqrt[4]{x} - 2\ln(\sqrt{x} + 1) + C$$

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