### Related Formula Standard power integration formula rule is: int u^n \, du = fracu^n+1n+1 + C$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ ### Core Logic Rewrite the integral by adjusting powers inside the radical container: I = int left(frac1x^2 + frac1x^4right) left(frac3x + frac1x^3right)^frac123 \, dx$I = \int \left(\frac{1}{x^2} + \frac{1}{x^4}\right) \left(\frac{3}{x} + \frac{1}{x^3}\right)^{\frac{1}{23}} \, dx$ Let t = frac3x + frac1x^3 implies dt = -3left(frac1x^2 + frac1x^4right)\,dx$t = \frac{3}{x} + \frac{1}{x^3} \implies dt = -3\left(\frac{1}{x^2} + \frac{1}{x^4}\right)\,dx$. ### Step 1: Integrate Substituting t$t$ into the equation: int fract^1/23 \, dt-3 = -frac13 cdot fract^24/2324/23 = -frac233(24) t^frac2423$\int \frac{t^{1/23} \, dt}{-3} = -\frac{1}{3} \cdot \frac{t^{24/23}}{24/23} = -\frac{23}{3(24)} t^{\frac{24}{23}}$ Comparing parameters directly yields: alpha = 23, quad beta = -1, quad gamma = -3$\alpha = 23, \quad \beta = -1, \quad \gamma = -3$ alpha + beta + gamma = 23 - 1 - 3 = 19$\alpha + \beta + \gamma = 23 - 1 - 3 = 19$ ### Pattern Recognition Pull out high powers from the root factor to seamlessly reveal a matching f'(x)$f'(x)$ substitution pair on the outside. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

### Related Formula Standard integration templates for quadratic forms: int sqrtt^2 + k^2 dt = fract2sqrtt^2+k^2 + frack^22lnleft|t + sqrtt^2+k^2right|$\int \sqrt{t^2 + k^2} dt = \frac{t}{2}\sqrt{t^2+k^2} + \frac{k^2}{2}\ln\left|t + \sqrt{t^2+k^2}\right|$ int frac1sqrtt^2 + k^2 dt = lnleft|t + sqrtt^2+k^2right|$\int \frac{1}{\sqrt{t^2 + k^2}} dt = \ln\left|t + \sqrt{t^2+k^2}\right|$ ### Core Logic Decompose the numerator using polynomial differentiation components : 2x^2 + 5x + 9 = A(x^2 + x + 1) + B(2x + 1) + C$2x^2 + 5x + 9 = A(x^2 + x + 1) + B(2x + 1) + C$ Equating coefficients dynamically yields : - For x^2$x^2$: A = 2$A = 2$ . - For x$x$: A + 2B = 5 Rightarrow 2 + 2B = 5 Rightarrow B = frac32$A + 2B = 5 \Rightarrow 2 + 2B = 5 \Rightarrow B = \frac{3}{2}$ . - Constant: A + B + C = 9 Rightarrow 2 + frac32 + C = 9 Rightarrow C = frac112$A + B + C = 9 \Rightarrow 2 + \frac{3}{2} + C = 9 \Rightarrow C = \frac{11}{2}$ . Rewrite the integrand into three parts : 2int sqrtx^2+x+1 dx + frac32int frac2x+1sqrtx^2+x+1 dx + frac112int frac1sqrtx^2+x+1 dx$2\int \sqrt{x^2+x+1} dx + \frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}} dx + \frac{11}{2}\int \frac{1}{\sqrt{x^2+x+1}} dx$ ### Step 1: Complete Quadratics & Integrate Format using completing the square technique: x^2 + x + 1 = left(x + frac12right)^2 + left(fracsqrt32right)^2$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$. 1. First Integral evaluation: 2 left[ fracleft(x+frac12right)2sqrtx^2+x+1 + frac38lnleft|x+frac12+sqrtx^2+x+1right| right]$2 \left[ \frac{\left(x+\frac{1}{2}\right)}{2}\sqrt{x^2+x+1} + \frac{3}{8}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| \right]$ = left(x+frac12right)sqrtx^2+x+1 + frac34lnleft|x+frac12+sqrtx^2+x+1right|$= \left(x+\frac{1}{2}\right)\sqrt{x^2+x+1} + \frac{3}{4}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$ 2. Second Integral evaluation : frac32 cdot 2sqrtx^2+x+1 = 3sqrtx^2+x+1$\frac{3}{2} \cdot 2\sqrt{x^2+x+1} = 3\sqrt{x^2+x+1}$ 3. Third Integral evaluation : frac112lnleft|x+frac12+sqrtx^2+x+1right|$\frac{11}{2}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$ ### Step 2: Collect Like Terms Gather and combine matching factor parameters: textTotal = left(x + frac12 + 3right)sqrtx^2+x+1 + left(frac34 + frac112right)lnleft|x+frac12+sqrtx^2+x+1right|$\text{Total} = \left(x + \frac{1}{2} + 3\right)\sqrt{x^2+x+1} + \left(\frac{3}{4} + \frac{11}{2}\right)\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$ textTotal = left(x + frac72right)sqrtx^2+x+1 + frac254lnleft|x+frac12+sqrtx^2+x+1right|$\text{Total} = \left(x + \frac{7}{2}\right)\sqrt{x^2+x+1} + \frac{25}{4}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$ = xsqrtx^2+x+1 + frac72sqrtx^2+x+1 + frac254lnleft|x+frac12+sqrtx^2+x+1right|$= x\sqrt{x^2+x+1} + \frac{7}{2}\sqrt{x^2+x+1} + \frac{25}{4}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$ ### Step 3: Extract Coefficients Compare directly against the given expression variables : alpha = frac72, quad beta = frac254$\alpha = \frac{7}{2}, \quad \beta = \frac{25}{4}$ alpha + 2beta = frac72 + 2left(frac254right) = frac72 + frac252 = frac322 = 16$\alpha + 2\beta = \frac{7}{2} + 2\left(\frac{25}{4}\right) = \frac{7}{2} + \frac{25}{2} = \frac{32}{2} = 16$ ### Pattern Recognition When dividing large numerators containing x^2$x^2$ elements over quadratic square roots, using matching coefficient expansion rules prevents lengthy substitution errors completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

### Related Formula Newton-Leibnitz Theorem for differentiation under integral sign: fracddx left( int_0^x tf(t) dt right) = xf(x)$\frac{d}{dx} \left( \int_{0}^{x} tf(t) dt \right) = xf(x)$ ### Core Logic Given: g(x) = int_0^x tf(t) dt$g(x) = \int_{0}^{x} tf(t) dt$ Differentiating both sides with respect to x$x$: g'(x) = xf(x) implies f(x) = fracg'(x)x$g'(x) = xf(x) \implies f(x) = \frac{g'(x)}{x}$ We are given g(x^3) = x^6 + x^7$g(x^3) = x^6 + x^7$. Let y = x^3 implies x = y^1/3$y = x^3 \implies x = y^{1/3}$. Substituting this into the expression for g$g$: g(y) = (y^1/3)^6 + (y^1/3)^7 = y^2 + y^7/3$g(y) = (y^{1/3})^6 + (y^{1/3})^7 = y^2 + y^{7/3}$ Thus, replacing y$y$ back with x$x$: g(x) = x^2 + x^7/3$g(x) = x^2 + x^{7/3}$ ### Step 1: Differentiate g(x) to find f(x) g'(x) = 2x + frac73x^4/3$g'(x) = 2x + \frac{7}{3}x^{4/3}$ Now find f(x)$f(x)$: f(x) = fracg'(x)x = frac2x + frac73x^4/3x = 2 + frac73x^1/3$f(x) = \frac{g'(x)}{x} = \frac{2x + \frac{7}{3}x^{4/3}}{x} = 2 + \frac{7}{3}x^{1/3}$ ### Step 2: Evaluate the Summation We need to find sum_r=1^15 f(r^3)$\sum_{r=1}^{15} f(r^3)$: f(r^3) = 2 + frac73(r^3)^1/3 = 2 + frac73r$f(r^3) = 2 + \frac{7}{3}(r^3)^{1/3} = 2 + \frac{7}{3}r$ Now, compute the summation from r=1$r=1$ to 15$15$: sum_r=1^15 f(r^3) = sum_r=1^15 left( 2 + frac73r right) = sum_r=1^15 2 + frac73sum_r=1^15 r$\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left( 2 + \frac{7}{3}r \right) = \sum_{r=1}^{15} 2 + \frac{7}{3}\sum_{r=1}^{15} r$ (2 times 15) + frac73 times frac15 times 162$(2 \times 15) + \frac{7}{3} \times \frac{15 \times 16}{2}$ 30 + frac73 times 120 = 30 + 7 times 40 = 30 + 280 = 310$30 + \frac{7}{3} \times 120 = 30 + 7 \times 40 = 30 + 280 = 310$ ### Pattern Recognition Converting g(x^3)$g(x^3)$ directly into a function of variable y=x^3$y=x^3$ prevents multi-layer chain rule complications when applying differentiation immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Definite Integration

### Related Formula Standard substitution method and logarithmic integral rule: int frac1t+1 dt = ln|t+1| + C$\int \frac{1}{t+1} dt = \ln|t+1| + C$ ### Core Logic Let x = t^4 implies dx = 4t^3 dt$x = t^4 \implies dx = 4t^3 dt$. When substituting into the integral: f(x) = int frac4t^3t(1+t) dt = 4 int fract^21+t dt$f(x) = \int \frac{4t^3}{t(1+t)} dt = 4 \int \frac{t^2}{1+t} dt$ ### Step 1: Simplify the Integral Rewrite the numerator t^2$t^2$ as (t^2 - 1) + 1$(t^2 - 1) + 1$: 4 int frac(t^2 - 1) + 11+t dt = 4 int left( frac(t-1)(t+1)1+t + frac11+t right) dt$4 \int \frac{(t^2 - 1) + 1}{1+t} dt = 4 \int \left( \frac{(t-1)(t+1)}{1+t} + \frac{1}{1+t} \right) dt$ 4 int (t - 1) dt + 4 int frac1t+1 dt$4 \int (t - 1) dt + 4 \int \frac{1}{t+1} dt$ 4 left[ frac(t-1)^22 right] + 4 ln|t+1| + C = 2(t-1)^2 + 4 ln|t+1| + C$4 \left[ \frac{(t-1)^2}{2} \right] + 4 \ln|t+1| + C = 2(t-1)^2 + 4 \ln|t+1| + C$ Substitute back t = x^1/4$t = x^{1/4}$: f(x) = 2(x^1/4 - 1)^2 + 4 ln(1 + x^1/4) + C$f(x) = 2(x^{1/4} - 1)^2 + 4 \ln(1 + x^{1/4}) + C$ ### Step 2: Solve for Constant C and Find f(1) Given f(0) = -6$f(0) = -6$: -6 = 2(0 - 1)^2 + 4 ln(1 + 0) + C$-6 = 2(0 - 1)^2 + 4 \ln(1 + 0) + C$ -6 = 2 + 0 + C implies C = -8$-6 = 2 + 0 + C \implies C = -8$ Now find f(1)$f(1)$: f(1) = 2(1^1/4 - 1)^2 + 4 ln(1 + 1^1/4) - 8$f(1) = 2(1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8$ f(1) = 2(0) + 4 ln(2) - 8 = 4 ln 2 - 8 = 4(ln 2 - 2)$f(1) = 2(0) + 4 \ln(2) - 8 = 4 \ln 2 - 8 = 4(\ln 2 - 2)$ ### Pattern Recognition By adding and subtracting terms in the numerator (t^2-1+1$t^2-1+1$), we can quickly bypass long division for polynomials and directly integrate using standard forms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Indefinite Integration

### Related Formula sin(x - theta) = sin x cos theta - cos x sin theta$\sin(x - \theta) = \sin x \cos \theta - \cos x \sin \theta$ ### Core Logic Let us partition the given integral I$I$ into two clean parts I_1$I_1$ and I_2$I_2$: I = int fracsin^3/2 xsqrtsin^3 x cos^3 x (sin x cos theta - cos x sin theta) \, dx + int fraccos^3/2 xsqrtsin^3 x cos^3 x (sin x cos theta - cos x sin theta) \, dx$I = \int \frac{\sin^{3/2} x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} \, dx + \int \frac{\cos^{3/2} x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} \, dx$ Factoring out appropriate powers of sin x$\sin x$ and cos x$\cos x$ from inside the roots: I_1 = int fracsec^2 xsqrttan x cos theta - sin theta \, dx$I_1 = \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} \, dx$ I_2 = int fraccsc^2 xsqrtcos theta - cot x sin theta \, dx$I_2 = \int \frac{\csc^2 x}{\sqrt{\cos \theta - \cot x \sin \theta}} \, dx$ ### Step 1: Evaluating the Integrals For I_1$I_1$, substitute tan x cos theta - sin theta = t^2 implies sec^2 x \, dx = frac2t \, dtcos theta$\tan x \cos \theta - \sin \theta = t^2 \implies \sec^2 x \, dx = \frac{2t \, dt}{\cos \theta}$: I_1 = int frac2t \, dtt cos theta = frac2tcos theta = 2sec theta sqrttan x cos theta - sin theta$I_1 = \int \frac{2t \, dt}{t \cos \theta} = \frac{2t}{\cos \theta} = 2\sec \theta \sqrt{\tan x \cos \theta - \sin \theta}$ For I_2$I_2$, substitute cos theta - cot x sin theta = z^2 implies csc^2 x \, dx = frac2z \, dzsin theta$\cos \theta - \cot x \sin \theta = z^2 \implies \csc^2 x \, dx = \frac{2z \, dz}{\sin \theta}$: I_2 = int frac2z \, dzz sin theta = frac2zsin theta = 2csc theta sqrtcos theta - cot x sin theta$I_2 = \int \frac{2z \, dz}{z \sin \theta} = \frac{2z}{\sin \theta} = 2\csc \theta \sqrt{\cos \theta - \cot x \sin \theta}$ ### Step 2: Combining Coefficients Comparing with the given expression, we find the coefficients: A = 2sec theta, quad B = 2csc theta$A = 2\sec \theta, \quad B = 2\csc \theta$ Multiplying them together: AB = 4sec theta csc theta = frac4sin theta cos theta = frac82sin theta cos theta = 8csc(2theta)$AB = 4\sec \theta \csc \theta = \frac{4}{\sin \theta \cos \theta} = \frac{8}{2\sin \theta \cos \theta} = 8\csc(2\theta)$ ### Pattern Recognition When integrating roots of trigonometric functions involving (x - theta)$(x - \theta)$, try dividing/multiplying fields by cos^n x$\cos^n x$ or sin^n x$\sin^n x$ to force tan x / sec^2 x$\tan x / \sec^2 x$ templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals