### 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