CBSE 2023 · Region 4 · Set 3 · Q26 · 3 marks
Find : \[\int \frac{2}{(1-x)\left(1+x^{2}\right)} d x \]
Marking-scheme solution
$$\text { Let } \frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{B x+C}{1+x^{2}}
\Rightarrow A=1, B=1, C=1
Hence, $\displaystyle \mathrm{I}=\int \frac{2}{(1-x)\left(1-x^{2}\right)} \mathrm{dx}=\int\left[\frac{1}{1-x}+\frac{x+1}{1+x^{2}}\right] \mathrm{dx}$
\begin{aligned}
& =\int \frac{1}{1-x} d x+\int \frac{x+1}{x^{2}+1} d x \\
& =\int \frac{1}{1-x} d x+\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x+\int \frac{1}{x^{2}+1} d x
\end{aligned}
=-\log |1-x|+\frac{1}{2} \log \left(x^{2}+1\right)+\tan ^{-1}(x)+C
$$
IntegralsIntegration by Partial FractionsApplyshort_answermedium
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 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.