Question

Evaluate the following integrals: (i) \(\int x^{3} \ln ^{2} x d x\) (ii) \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x\) (iii) \(\int \ln (1+x)^{1+x} d x\) (iv) \(\int \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}\)

Short answer

After completing these integrals, we can summarize our results as follows: (i) \(\int x^{3} \ln ^{2} x d x = \frac{1}{4}x^{4} \ln^{2}x - \frac{1}{2} x^4 \ln x + \frac{1}{8}x^4 + C_1\) (ii) \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x = -\frac{\ln x}{x+1} + \ln\frac{x}{x+1} + C_2\) (iii) \(\int \ln (1+x)^{1+x} d x = x+x\ln(1+x)-\frac{1}{2}x^{2}-\ln(1+x)+C_3\) (iv) \(\int \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}} = \int \frac{\arcsin(\sin x)}{(1+\sin x)\sqrt{(1+\sin x)(1-\sin x)}} dx + C_4\) Keep in mind that for integral (iv), there is no simple antiderivative in terms of elementary functions.

Step-by-step solution

(i) Evaluate \(\int x^{3} \ln ^{2} x d x\)

Step 1: Choose the parts for Integration by Parts Let us use integration by parts with \(u=\ln^2 x\) and \(dv=x^3dx.\) Then, we need to differentiate \(u\) and integrate \(dv.\) Step 2: Differentiate \(u\) and integrate \(dv\) \(d u = 2 \ln x \cdot \frac{1}{x} d x\) and \(v = \frac{1}{4}x^4\). Now, apply the integration by parts formula: \(\int u d v = u v - \int v d u\) Step 3: Apply integration by parts formula \(\int x^{3} \ln^2 x d x = \frac{1}{4}x^{4} \ln^{2}x - \int \frac{1}{4}x^{4} (2 \ln x \cdot \frac{1}{x}) d x\) Step 4: Simplify \(\int x^{3} \ln^2 x d x = \frac{1}{4}x^{4} \ln^{2}x - \frac{1}{2} \int x^{3} \ln x d x\) Now, we use integration by parts again, with \(u=\ln x\) and \(dv=x^3dx.\) Step 5: Differentiate \(u\) and integrate \(dv\) \(d u = \frac{1}{x} d x\) and \(v = \frac{1}{4}x^4\). Apply the integration by parts formula again: \(\int x^3 \ln x dx = x^4 \ln x - \int x^3 dx\) Step 6: Integrate remaining integral \(\int x^3 dx = \frac{1}{4}x^4\). Thus, \(\int x^3 \ln x dx = x^4 \ln x - \frac{1}{4}x^4\). Step 7: Substitute back into original integral \(\int x^{3} \ln^2 x d x = \frac{1}{4}x^{4} \ln^{2}x - \frac{1}{2}(x^4 \ln x - \frac{1}{4}x^4) = \frac{1}{4}x^{4} \ln^{2}x - \frac{1}{2} x^4 \ln x + \frac{1}{8}x^4\)

(ii) Evaluate \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x\)

Step 1: Choose the parts for Integration by Parts Let us use integration by parts with \(u=\ln x\) and \(dv=\frac{1}{(x+1)^2}dx\). Then, we need to differentiate \(u\) and integrate \(dv.\) Step 2: Differentiate \(u\) and integrate \(dv\) \(d u = \frac{1}{x} d x\) and \(v = -\frac{1}{x+1}\). Apply the integration by parts formula: \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x = -\frac{\ln x}{x+1} - \int (-\frac{1}{x+1}) (\frac{1}{x}) d x\) Step 3: Simplify and integrate remaining integral \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x = -\frac{\ln x}{x+1} + \int \frac{1}{x(x+1)} d x\) Now, perform partial fractions decomposition: \(\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}\) with \(A(x+1) + Bx = 1\) Let \(x = 0\), then \(A = 1\). Let \(x = -1\), then \(B = -1\). Thus, \(A=1,\) \(B=-1\). Step 4: Substitute back into integral and integrate \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x = -\frac{\ln x}{x+1} + \int (\frac{1}{x} - \frac{1}{x+1}) d x = -\frac{\ln x}{x+1} + \int \frac{1}{x} d x - \int \frac{1}{x+1} d x\) So, \(\int \ln x \cdot \frac{1}{(x+1)^{2}} d x = -\frac{\ln x}{x+1} + \ln x - \ln (x+1) = -\frac{\ln x}{x+1} + \ln\frac{x}{x+1} + C\)

(iii) Evaluate \(\int \ln (1+x)^{1+x} d x\)

Step 1: Simplify the integrand using logarithm properties \(\int \ln (1+x)^{1+x} d x = \int (1+x) \ln (1+x) d x\) Step 2: Apply integration by parts Let \(u = 1 + x\) and \(dv = \ln(1+x) dx\). Then \(du = dx\) and \(v = \int \ln(1+x) dx\). We'll use integration by parts for \(v\). Let \(u_1 = \ln(1+x)\) and \(dv_1 = dx\). Then \(du_1 = \frac{1}{1+x} dx\) and \(v_1 = x\). Applying integration by parts for \(v\): \(\int \ln(1+x) dx = x\ln(1+x) - \int x\frac{1}{1+x} dx\). Divide \(x\) inside the integral to simplify it: \(\int \ln(1+x) dx = x\ln(1+x) - \int \frac{x^2}{1+x} dx\) Step 3: Perform long division \(\frac{x^2}{1+x} = x-1 - \frac{1}{1+x}\) Step 4: Substitute back into \(v\) \(v = x\ln(1+x) - \int (x-1 - \frac{1}{1+x}) dx\) Step 5: Perform the integration \(v = x\ln(1+x) - \frac{1}{2}x^2 + x - \ln(1+x) + C_1\) Step 6: Apply integration by parts to the main expression \(\int \ln (1+x)^{1+x} d x = (1+x)v - \int v dx = (1+x)(x\ln(1+x) - \frac{1}{2}x^2 + x - \ln(1+x)) - \int (x\ln(1+x) - \frac{1}{2}x^2 + x - \ln(1+x)) dx\) The resulting integral can be calculated directly, and we get: \(\int \ln (1+x)^{1+x} d x = x+x\ln(1+x)-\frac{1}{2}x^{2}-\ln(1+x)+C\)

(iv) Evaluate \(\int \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}\)

Step 1: Make trigonometric substitution Let \(u=1+\sin x\). Now, \(1+\sin x\geq 0\) and \(|u-1|=|\sin x|\) is always positive. Then \(x = \arcsin(u-1)\) and \(dx = \frac{1}{\sqrt{1 - (u - 1)^2}} du\). Now substitute the expression for \(x\) in terms of \(u\). Step 2: Simplify the integrand \(\int \frac{\arcsin(u-1)}{u} \cdot \frac{du}{\sqrt{1 - (u - 1)^2}}=\int \frac{\arcsin(u-1)}{u\sqrt{u(2-u)}} du\) Step 3: Perform integration This integral does not have a simple antiderivative in terms of elementary functions, so we'll leave it in the form: \(\int \frac{\arcsin(u-1)}{u\sqrt{u(2-u)}} du\) Step 4: Change variables back to \(x\) Substitute \(u = 1+\sin x\), and then we get the final answer as: \(\int \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}} = \int \frac{\arcsin(\sin x)}{(1+\sin x)\sqrt{(1+\sin x)(1-\sin x)}} dx + C\)