Q.Evaluate ∫x2+6x+13x+3dx.
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Ten named integral patterns -- ∫dx/(x2±a2), ∫dx/x2±a2, ∫dx/a2−x2, and their generalisations to a full quadratic ax2+bx+c under a denominator
or a square root (reduced via completing the square), plus the linear-numerator forms ∫(px+q)dx/(ax2+bx+c) and ∫(px+q)dx/ax2+bx+c (numerator split into a multiple of …
Numerator x+3 is exactly half the derivative of the radicand. …
The radicand's derivative is 2x+6=2(x+3), exactly twice the numerator. Let u=x2+6x+13,
du=(2x+6)dx=2(x+3)dx, so (x+3)dx=21du:
∫x2+6x+13x+3dx=21∫u−1/2du=21⋅2u+C=x2+6x+13+C. …
Notice the numerator is an exact constant multiple of the derivative of the radicand; substitute u= the radicand directly (no numerator-splitting remainde …
Missing that the numerator matches the radicand's derivative up to a constant, and instead attempting an unnecessary full split into …
Showing the 12 most recent of 30 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.∫ (from 0 to 1) dx/√(1-x²) is equal to:(a) π/2(b) π/4(c) π/3(d) π/6
›Reveal solutionSolution
1−x21 is the derivative of sin−1x, so the definite integral is just sin−1(1)−sin−1(0).
We know the standard result ∫1−x2dx=sin−1x+C.
So: …
- CBSE 2025Set ANNUAL1 markMCQQ.If ∫2xtt2−1dt=12π, then the value of x is(a) 2(b) 1(c) 0(d) -1
›Reveal solutionSolution
The antiderivative of tt2−11 is sec−1t; apply the limits and solve for x.
We know the standard result:
∫tt2−1dt=sec−1∣t∣+C
So
∫2xtt2−1dt=sec−1x−sec−12
Since sec4π=2, we have sec−12=4π.
Given this equals 12π:
sec−1x−4π=12π …
- CBSE 2024Set A1 markQ.Match the Correct Columns. Column A: ∫x2−a2dx. Column B options:(a) 2a1logx+ax−a+c(b) sin−1ax+c(c) 2xa2−x2+2a2sin−1ax+c(d) logx+x2−a2+c(e) 2xx2−a2−2a2logx+x2−a2+c(f) sec−1x(g) 2a1loga−xa+x+c(h) a1tan−1ax+c(i) tan−1x. Which entry of Column B matches this Column A item?
›Reveal solutionSolution
∫x2−a2dx matches entry (e), the standard NCERT formula for this integral.
…
- CBSE 2024Set A1 markQ.Match the Correct Columns. Column A: ∫a2−x2dx. Column B options:(a) 2a1logx+ax−a+c(b) sin−1ax+c(c) 2xa2−x2+2a2sin−1ax+c(d) logx+x2−a2+c(e) 2xx2−a2−2a2logx+x2−a2+c(f) sec−1x(g) 2a1loga−xa+x+c(h) a1tan−1ax+c(i) tan−1x. Which entry of Column B matches this Column A item?
›Reveal solutionSolution
∫a2−x2dx matches entry (c), the standard NCERT formula for this integral.
…
- CBSE 2024Set A1 markQ.Match the Correct Columns. Column A: ∫x2−a2dx. Column B options:(a) 2a1logx+ax−a+c(b) sin−1ax+c(c) 2xa2−x2+2a2sin−1ax+c(d) logx+x2−a2+c(e) 2xx2−a2−2a2logx+x2−a2+c(f) sec−1x(g) 2a1loga−xa+x+c(h) a1tan−1ax+c(i) tan−1x. Which entry of Column B matches this Column A item?
›Reveal solutionSolution
∫x2−a2dx matches entry (d).
…
- CBSE 2024Set A1 markQ.Match the Correct Columns. Column A: ∫a2−x2dx. Column B options:(a) 2a1logx+ax−a+c(b) sin−1ax+c(c) 2xa2−x2+2a2sin−1ax+c(d) logx+x2−a2+c(e) 2xx2−a2−2a2logx+x2−a2+c(f) sec−1x(g) 2a1loga−xa+x+c(h) a1tan−1ax+c(i) tan−1x. Which entry of Column B matches this Column A item?
›Reveal solutionSolution
∫a2−x2dx matches entry (b).
…
- CBSE 2024Set A1 markQ.Match the Correct Columns. Column A: ∫x2−a2dx. Column B options:(a) 2a1logx+ax−a+c(b) sin−1ax+c(c) 2xa2−x2+2a2sin−1ax+c(d) logx+x2−a2+c(e) 2xx2−a2−2a2logx+x2−a2+c(f) sec−1x(g) 2a1loga−xa+x+c(h) a1tan−1ax+c(i) tan−1x. Which entry of Column B matches this Column A item?
›Reveal solutionSolution
∫x2−a2dx matches entry (a).
…
- CBSE 2024Set A1 markQ.Match the Correct Columns. Column A: ∫a2−x2dx. Column B options:(a) 2a1logx+ax−a+c(b) sin−1ax+c(c) 2xa2−x2+2a2sin−1ax+c(d) logx+x2−a2+c(e) 2xx2−a2−2a2logx+x2−a2+c(f) sec−1x(g) 2a1loga−xa+x+c(h) a1tan−1ax+c(i) tan−1x. Which entry of Column B matches this Column A item?
›Reveal solutionSolution
∫a2−x2dx matches entry (g).
…
- CBSE 2023Set A1 markQ.Match the following — find the correct match for ∫tanxdx from:(a) log∣sinx∣+c(b) cos2x(c) sec2x(d) 2cos2x(e) log∣cosecx−cotx∣+c(f) −log∣cosx∣+c(g) log∣secx+tanx∣+c
›Reveal solutionSolution
∫tanxdx=−log∣cosx∣+c, matching option (f).
Write tanx=cosxsinx and substitute u=cosx, du=−sinxdx: …
- CBSE 2023Set A1 markQ.Match the following — find the correct match for ∫cotxdx from:(a) log∣sinx∣+c(b) cos2x(c) sec2x(d) 2cos2x(e) log∣cosecx−cotx∣+c(f) −log∣cosx∣+c(g) log∣secx+tanx∣+c
›Reveal solutionSolution
∫cotxdx=log∣sinx∣+c, matching option (a).
Write cotx=sinxcosx and substitute u=sinx, du=cosxdx: …
- CBSE 2023Set A1 markQ.Match the following — find the correct match for ∫secxdx from:(a) log∣sinx∣+c(b) cos2x(c) sec2x(d) 2cos2x(e) log∣cosecx−cotx∣+c(f) −log∣cosx∣+c(g) log∣secx+tanx∣+c
›Reveal solutionSolution
∫secxdx=log∣secx+tanx∣+c, matching option (g). This is a standard result — a memorised integral, verified by differentiating the RHS.
…
- CBSE 2023Set A1 markQ.Match the following — find the correct match for ∫cosecxdx from:(a) log∣sinx∣+c(b) cos2x(c) sec2x(d) 2cos2x(e) log∣cosecx−cotx∣+c(f) −log∣cosx∣+c(g) log∣secx+tanx∣+c
›Reveal solutionSolution
∫cosecxdx=log∣cosecx−cotx∣+c, matching option (e). A standard integral, verified by differentiating the RHS.
…
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