CBSE 2026 · Region 3 · Set 1 · Q26 · 3 marks
Find : \[\int \frac{d \mathrm{x}}{\mathrm{x}^{1 / 2}+\mathrm{x}^{1 / 3}} \]Find : \[\int \tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right) d \mathrm{x} \]
Find : \[\int \frac{d \mathrm{x}}{\mathrm{x}^{1 / 2}+\mathrm{x}^{1 / 3}} \]
Find : \[\int \tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right) d \mathrm{x} \]
Marking-scheme solution
Put $\displaystyle x=t^{6} \Rightarrow dx=6 t^{5} dt$
$\displaystyle \int \dfrac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=6 \int \dfrac{t^{5}}{t^{3}+t^{2}} dt=6 \int \dfrac{t^{3}}{t+1} dt$
$\displaystyle =6 \int\left(t^{2}-t+1-\dfrac{1}{t+1}\right) dt$
$\displaystyle =6\left(\dfrac{t^{3}}{3}-\dfrac{t^{2}}{2}+t-\log |t+1|\right)+c$
$\displaystyle =2 x^{\frac{1}{2}}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log\left|x^{\frac{1}{6}}+1\right|+c$
$\displaystyle \int \tan^{-1}\left(\dfrac{1-x}{1+x}\right) dx$
Getting $\displaystyle \int\left(\dfrac{\pi}{4}-\tan^{-1} x\right) dx$
$\displaystyle =\dfrac{\pi x}{4}-\left[x \tan^{-1} x-\int \dfrac{x}{1+x^{2}} dx\right]$
$\displaystyle =\dfrac{\pi x}{4}-x \tan^{-1} x+\dfrac{1}{2} \log\left(1+x^{2}\right)+c$
IntegralsMethods of IntegrationApplyshort_answerhard
More from Integrals
- ∫ (x+5)/((x+6)^2) e^x dx is equal to:2025 · asked 3×
- Evaluate ∫ 1/((e^x+e^-x)(e^x-e^-x)) d x log √22023 · asked 3×
- If ∫ (3 a x)/( b^2+c^2 x^2) d x=A log b^2+c^2 x^2 +K, then the value of A is2026 · asked 3×
- Find: Find: ∫ x^2 log (x^2+1) d x2023 · asked 3×
- If ∫ (2^1/x)/(x^2) d x=k · 2^1/x+C, then k is equal to2025 · asked 3×
- Find: ∫ (x^2+1)/((x-1)^2(x+3)) d x OR Evaluate: ∫ 0^π / 2 (x)/( sin x+ cos x) d x2025 · asked 3×
- Find: ∫ dx√4 x-x^22022 · asked 3×
- Find: ∫ (2 x)/((x^2+3)(x^2-5)) d x OR Evaluate: ∫ 1^4( x-2 + x-4 ) d x2025 · asked 3×
CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.